Vertex isoperimetry and independent set stability for tensor powers of cliques

TL;DR

This paper investigates vertex isoperimetry and the stability of independent sets in tensor powers of cliques, providing recursive formulas and improved bounds that have implications for graph coloring and approximation hardness.

Contribution

It introduces a recursive method to compute vertex isoperimetric bounds and establishes near-optimal stability bounds for independent sets in tensor powers of cliques.

Findings

Recursive relation for vertex isoperimetry function ()

Improved stability bounds for independent sets with exponent _t()

Results applicable to hardness of approximation in graph problems

Abstract

The tensor power of the clique on $t$ vertices (denoted by $K_t^n$) is the graph on vertex set $\{1, ..., t\}^n$ such that two vertices $x, y \in \{1, ..., t\}^n$ are connected if and only if $x_i \neq y_i$ for all $i \in \{1, ..., n\}$. Let the density of a subset $S$ of $K_t^n$ to be $\mu(S) := \frac{|S|}{t^n}$, and let the vertex boundary of a set $S$ to be vertices which are incident to some vertex of $S$, perhaps including points of $S$. We investigate two similar problems on such graphs. First, we study the vertex isoperimetry problem. Given a density $\nu \in [0, 1]$ what is the smallest possible density of the vertex boundary of a subset of $K_t^n$ of density $\nu$? Let $\Phi_t(\nu)$ be the infimum of these minimum densities as $n \to \infty$. We find a recursive relation allows one to compute $\Phi_t(\nu)$ in time polynomial to the number of desired bits of precision. Second, we study given an independent set $I \subseteq K_t^n$ of density $\mu(I) = \frac{1}{t}(1-\epsilon)$, how close it is to a maximum-sized independent set $J$ of density $\frac{1}{t}$. We show that this deviation (measured by $\mu(I \setminus J)$) is at most $4\epsilon^{\frac{\log t}{\log t - \log(t-1)}}$ as long as $\epsilon < 1 - \frac{3}{t} + \frac{2}{t^2}$. This substantially improves on results of Alon, Dinur, Friedgut, and Sudakov (2004) and Ghandehari and Hatami (2008) which had an $O(\epsilon)$ upper bound. We also show the exponent $\frac{\log t}{\log t - \log(t-1)}$ is optimal assuming $n$ tending to infinity and $\epsilon$ tending to $0$. The methods have similarity to recent work by Ellis, Keller, and Lifshitz (2016) in the context of Kneser graphs and other settings. The author hopes that these results have potential applications in hardness of approximation, particularly in approximate graph coloring and independent set problems.

Equations

$$\partial S:=\{x\in V_G:\text{exists y∈S such that {x,y}∈EG}\}.$$

$$\mu (S):=\frac{|S|}{|V_G|}.$$

$$\Phi (G,\nu )=\inf \{\mu (\partial S):\mu (S)\ge \nu \}$$

$$h(G)$$

$$\eta (G,ϵ)$$

$$\eta (K_t^n)=\eta (K_t)=\frac{\log t}{\log t-\log (t-1)}=t\log t+\Theta (\log t).$$

$$J=[t{]}^{i-1}\times \{j\}\times [t{]}^{n-i}.$$

$$\mu (I\setminus J)\le 4{ϵ}^{\eta (K_t)}=4{ϵ}^{\log t/(\log t-\log (t-1))}.$$

$$\mu (I\Delta J)=\mu (I\setminus J)+\mu (J\setminus I)=\mu (J)-\mu (I)+2\mu (I\setminus J)=\frac{ϵ}{t}+4{ϵ}^{\eta (K_t)},$$

$$I=\{(1,1,a),(1,a,1),(a,1,1):a\in [t]\}\times [t{]}^{n-3}.$$

$$c_i(S)=\left\{x\in [t{]}^n:x_i\le |\{y\in S:y_{\neg i}=x_{\neg i}\}|\right\}.$$

$$\Sigma (S):=\sum _{x\in S}\sum _{j\in [n]}{x}_j$$

$$x_{\neg i}$$

$$y_{\neg i}=z_{\neg i}$$

$$y_i^{\prime }$$

$$\eta (t)$$

$$\eta (K_t^n)\le \frac{\log \mu (S)}{\log \mu (\partial S)}=\frac{\log t^{-k}}{\log ((t-1{)}^k{t}^{-k})}=\frac{k\log \frac{1}{t}}{k\log \frac{t-1}{t}}=\eta (t).\text{qed}$$

$$y^{1/\eta (t)}+(t-1)x^{1/\eta (t)}\ge (t-1)(x+(t-1)y{)}^{1/\eta (t)}$$

$$\mu (\partial (S))\ge \mu (S{)}^{1/\eta (t)}.$$

$$S_i$$

$$(\partial S{)}_i$$

$$\mu (\partial S)$$

$$\ge \frac{1}{t}(\mu (\partial S_1)+(t-1)\mu (\partial S_0))$$

$$\ge \frac{1}{t}\left(\mu (S_1{)}^{1/\eta (t)}+(t-1)\mu (S_0{)}^{1/\eta (t)}\right),$$

$$\mu (\partial (S))$$

$$\ge \frac{t-1}{t}{\left(\mu (S_0)+(t-1)\mu (S_1)\right)}^{1/\eta (t)}$$

$$\ge \frac{t-1}{t}{\left(\sum _{i\in [t]}\mu (S_i)\right)}^{1/\eta (t)}$$

$$={\left(\frac{1}{t}\sum _{i\in [t]}\mu (S_i)\right)}^{1/\eta (t)}$$

$$=\mu (S{)}^{1/\eta (t)},$$

$$\Phi (G,\nu ):=\inf \{\mu (\partial S):\mu (S)\ge \nu \}.$$

$${\Phi }_t(\nu ):=\underset{n\ge 1}{\inf }\Phi (K_t^n,\nu ).$$

$${\Phi }_t(\nu )\ge {\nu }^{1/\eta (t)}.$$

$${\Phi }_t(\nu )=\{\begin{array}{ll}\frac{t-1}{t}{\Phi }_t(t\nu )& \nu <1/t\\ \frac{t-1}{t}+\frac{1}{t}{\Phi }_t\left(\frac{t\nu -1}{t-1}\right)& \nu \ge 1/t\end{array}.$$

$${\Phi }_3\left(\frac{5}{9}\right)=\frac{2}{3}+\frac{1}{3}{\Phi }_3\left(\frac{1}{3}\right)=\frac{8}{9},$$

$$\mu (I\setminus J)<\frac{1}{t^3}.$$

Full text

First, we study the vertex isoperimetry problem. Given a density $\nu\in[0,1]$ what is the smallest possible density of the vertex boundary of a subset of $K_{t}^{n}$ of density $\nu$? Let $\Phi_{t}(\nu)$ be the infimum of these minimum densities as $n\to\infty$. We find a recursive relation allows one to compute $\Phi_{t}(\nu)$ in time polynomial to the number of desired bits of precision.

Second, we study given an independent set $I\subseteq K_{t}^{n}$ of density $\mu(I)=\frac{1}{t}(1-\epsilon)$, how close it is to a maximum-sized independent set $J$ of density $\frac{1}{t}$. We show that this deviation (measured by $\mu(I\setminus J)$) is at most $4\epsilon^{\frac{\log t}{\log t-\log(t-1)}}$ as long as $\epsilon<1-\frac{3}{t}+\frac{2}{t^{2}}$. This substantially improves on results of Alon, Dinur, Friedgut, and Sudakov (2004) and Ghandehari and Hatami (2008) which had an $O(\epsilon)$ upper bound. We also show the exponent $\frac{\log t}{\log t-\log(t-1)}$ is optimal assuming $n$ tending to infinity and $\epsilon$ tending to [math]. The methods have similarity to recent work by Ellis, Keller, and Lifshitz (2016) in the context of Kneser graphs and other settings.

The author hopes that these results have potential applications in hardness of approximation, particularly in approximate graph coloring and independent set problems.

Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213, USA. Email: jbrakens at andrew.cmu.edu. This work was partially supported by REU supplements to NSF CCF-1526092 and CCF-1422045

1. Introduction

1.1. Vertex isoperimetry

For any undirected graph $G=(V_{G},E_{G})$ and $S\subseteq V_{G}$, we define the vertex boundary of $S$ to be

[TABLE]

Furthermore, we define the density of $S$ to be

[TABLE]

The relationship between $\mu(S)$ and $\mu(\partial S)$, particularly when $\mu(S)$ is sufficiently small (typically at most $1/2$). Is known as a vertex isoperimetric inequality. Such relationships are captured by the isoperimetric parameter (or isoperimetric profile) of a graph

[TABLE]

Proving such inequalities for various graphs is a frequent topic in the literature (e.g., [BHT00, CEK13]). Typically such works focus on a linear or near-linear relationship between $\mu(\partial S)$ and $\mu(S)$, known as the isoperimetric constant.

[TABLE]

In this paper, we study graphs for which there is an order-of-magnitude difference between $\mu(S)$ and $\mu(\partial S)$, when $\mu(S)$ is sufficiently small. For example, if $\mu(\partial S)\geq\sqrt{\mu(\partial S)}$ for all $S$, we would like to say that $G$ expands by a power of $2$. Such ‘hyper-expansion’ can be captured by what we coin as the isoperimetric exponent. For all $\epsilon>0$ consider.

[TABLE]

where $\log$ is the natural logarithm. In other words, for every subset $S$ of $G$ of density $\delta$, the boundary of $S$ has density at least $\delta^{1/\eta(G,\epsilon)}.$ The larger the parameter $\eta(G)$ is, the more ‘expansive’ the graph is. It is easy to see that $\eta(G,\epsilon)$ is in general a decreasing function of $\epsilon$. As we often work with large subsets of our graph, we let $\eta(G):=\eta(G,1)$.

In this paper, we study the isoperimetric profile of the tensor powers of cliques. For undirected graphs $G=(V_{G},E_{G}),H=(V_{H},E_{H})$, we define the tensor product $G\otimes H$ to be the undirected graph on vertex set $V_{1}\times V_{2}$ such that an edge connects $(u_{1},v_{1})$ and $(u_{2},v_{2})$ if and only if $\{u_{1},u_{2}\}\in E_{G}$, and $\{v_{1},v_{2}\}\in E_{H}.$ Note that up to isomorphism, the tensor product is both commutative and associative. We then denote $\otimes^{n}G$ to be the tensor product of $n$ copies of $G$. Since this is the only graph product discussed in this article, we shorten this to $G^{n}$. In this article, we focus on the case that $G=K_{t}$, where $K_{t}$ is the complete graph on $t\geq 3$ vertices. It turns out for such graphs that for all $\epsilon>\frac{1}{t^{n}}$, $\eta(G)=\eta(G,\epsilon)$.

In particular, we shall compute the following.

Theorem 1**.**

For all $t\geq 3$ and all positive integers $n$,

[TABLE]

In addition to this high-level structure, we give a more-fine-tuned analysis of the behavior of $\Phi_{t}(\eta):=\inf_{n\geq 1}\Phi(K_{t}^{n},\eta)$. (See Theorem 10.)

1.2. Independent set stability

With these vertex isoperimetric inequalities, we apply them to the understanding the structure of near-maximum independent sets of graphs. Such results are known as stability results.

Such results are not just of interest within combinatorics, a better understanding of independent set stability of certain graphs, such as $K_{t}^{n}$, have resulted in advances in hardness of approximation, particularly in construct dictatorship tests for approximate graph coloring and independent set problems (e.g., [ADFS04, DFR08, BG16]). In fact the investigation which led to the results in this paper was inspired by the pursuit of such results.

A landmark result of this form due to [ADFS04] is as follows.

Theorem 2** ([ADFS04]).**

For all $t\geq 3$ there exist $C_{t}$ with the following property. For any positive integer $n$, Let $I\subset\ K_{t}^{n}$ be an independent set such that $\epsilon=1-t\mu(I)$, then there exists an independent set $J\subset K_{t}^{n}$ of maximum size ($\mu(J)=1/t$) such that $\mu(I\Delta J)\leq C_{t}\epsilon$, where $S\Delta T=(S\setminus T)\cup(T\setminus S)$.

In other words, independent sets of near-maximum size are similar in structure to the maximum independent sets. Note that if $J$ is an independent set of maximum size, then for some $i\in[n]$ and $j\in[t]$, we have that

[TABLE]

This is a well-known result due to [GL74] (see [AS04] for a proof using Fourier analysis).

Ghandehari and Hatami improved this result (Theorem 1 of [GH08]) to show that if $t\geq 20$ and $\epsilon\leq 10^{-9}$ then $C_{t}$ can be replaced with $40/t$. Both results were proven using Fourier analysis.

We improve upon this result in two steps. First, with an application of Theorem 1 we improve Theorem 2 in a black-box matter to obtain.

Theorem 3**.**

For all $t\geq 3$, there exists $\epsilon_{t}>0$ with the following property. For any positive integer $n$, Let $I\subset\ K_{t}^{n}$ be an independent set such that $\epsilon=1-t\mu(I)<\epsilon_{t}$, then there exists an independent set $J\subset K_{t}^{n}$ of maximum size ($\mu(J)=1/t$) such that

[TABLE]

Remark 1*.*

Since $\mu(I\setminus J)\leq 4\epsilon^{\eta(K_{t})}$,

[TABLE]

so our result gives the optimal first-order structure for Theorem 2 assuming $\epsilon$ is sufficiently small. Furthermore, in Appendix C, we give examples of independent sets of $K_{t}^{n}$ with arbitrarily small density (assuming $n\to\infty$) for which the exponent $\eta(K_{t})$ is optimal.

Next, using a purely combinatorial argument we pin down a precise value for $\epsilon_{t}$.

Theorem 4**.**

In Theorem 3, for all $t\geq 3$, one may set $\epsilon_{t}=1-\frac{3}{t}+\frac{2}{t^{2}}$. In other words, the theorem applies for all independent sets $I$ such that $\mu(I)>\frac{3t-2}{t^{3}}$.

The choice of $\epsilon_{t}$ is not arbitrary, it corresponds to the density of the following independent set.

[TABLE]

Note that $\mu(I)=\frac{3t-2}{t^{3}}.$ This set represents a phase transition in the independent sets from ‘dictators’ to ‘juntas,’ as the $I$ constructed above is equally influenced by $3$ coordinates (where ‘influence’ is in the sense of [ADFS04]). Such phase transitions have been studied in the literature [DFR08], but this may be the first work to highlight the exact transition point.

Additionally, to the best of the author’s knowledge, this is the first known purely combinatorial proof of Theorem 2.

1.2.1. Related work

Such stability results for independent sets have also been studied for Kneser graphs. A result similar to that of Theorem 2 was proved by [Fri08]. Numerous other works in the literature [DF09, DS05, BM08, Kee08, KM10, FKMW16, FM16] prove generalized stability results for Kneser graphs or other structures related to intersecting families.

A result which also finds a “tight” super constant exponent $\eta>1$ for the independent set stability is proved in some very recent work [EKL16b, EKL16a, EL16, KL16b, KL16a, EKL17] on Kneser graphs and related structures. (See also [EKN17] and Proposition 4.3 of [Fil16].) The techniques have high-level similarity to the ones adopted here:111The author became aware of these similar proofs only after writing major portions of the manuscript. particularly in their use of compressions to prove a isoperimetric inequality which they then bootstrap to a combinatorial independent set stability result.

1.3. Paper organization

In Section 2 we prove the claimed vertex isoperimetric inequalities. In Section 3, we prove the stability results for near-maximum independent sets in $K_{t}^{n}$. Appendix A proves some algebraic inequalities omitted from the main text. Appendix B proves Theorem 10, which gives a refined understanding the isoperimetric profile of Kneser graphs. Appendix C shows that the exponent of $\eta(t)$ in Theorems 3 and 4 is optimal.

2. Vertex isoperimetric Inequalities

In this section, we proceed to prove the isoperimetry results claimed in Section 1.1.

Identify the vertex set of $K_{t}^{n}$ with $[t]^{n}$. Two vertices of $x,y\in[t]^{n}$ are connected in $K_{t}^{n}$ if and only if $x_{i}\neq y_{i}$ for all $i\in[n].$ Denote $y_{\neg i}:=(y_{1},\ldots,y_{i-1},y_{i+1},\ldots,y_{n})$. We often write $y$ as $(y_{i},y_{\neg i})$ when it is clear from context which coordinate is being inserted.

2.1. Compressions

A useful tool in our study will be the operation of the well-known technique of compressions (e.g., [Sau72, She72]). Although compressions are not strictly necessary to prove Theorem 1, they are essential in the proof of stronger isoperimetry results as well as Theorem 4, so we introduce the machinery now.

For $S\subseteq[t]^{n}$ be a subset, define the compression of $S$ in coordinate $i$ to be

[TABLE]

Informally, we ‘shift’ each element of $S$ to be as small as possible in the $i$th direction. Note that $\mu(c_{i}(S))=\mu(S)$ for all $S\subseteq\mathbb{[}t]^{n}$. It is easy to see that $c_{i}$ is nilpotent: $c_{i}(c_{i}(S))=c_{i}(S)$ for all $S\subseteq[t]^{n}$ and $i\in[n]$.

We say that a set $S$ is compressed if $c_{i}(S)=S$ for all $i\in[n]$. Equivalently, for all $x\in S$ there is no $y\in[t]^{n}\setminus S$ such that $x_{i}\leq y_{i}$ for all $i\in[n]$.

Remark 2*.*

Note that every time a compression $c_{i}$ is applied, the quantity

[TABLE]

decreases or stays the same (in which case $c_{i}(S)=S$). Thus, since $\Sigma(S)$ is always positive, there must exist a finite sequence of compressions which can be applied to $S$ to make the set compressed.

Now we show that compressions respect independent sets of $K_{t}^{n}$. This result is not needed until Section 3, but the proof does give intuition for how the compressions work.

Claim 5**.**

For all $i\in[n]$ and all $I\subset\mathbb{[}t]^{n}$ independent set of $K_{t}^{n}$, $c_{i}(I)$ is also an independent set of $K_{t}^{n}$.

Proof.

Assume not, then there exist $x,y\in c_{i}(I)$ such that $\{x,y\}$ is an edge. In particular, since $x_{i}\neq y_{i}$, we must have that $x_{i}\neq 1$ or $y_{i}\neq 1$. Assume without loss of generality that $y_{i}\neq 1$. Then, by definition of $c_{i}(I)$, there must be $z:=(1,y_{\neg i})\in c_{i}(I)$. Since $x,y,z\in c_{i}(I)$, there must be $x^{\prime},y^{\prime},z^{\prime}\in I$ such that

[TABLE]

Since $y_{i}^{\prime}\neq z^{\prime}_{i}$, we must either have that $x_{i}^{\prime}\neq y_{i}^{\prime}$ or $x_{i}^{\prime}\neq z_{i}^{\prime}$. In the former case, $\{x^{\prime},y^{\prime}\}$ is an edge of $K_{t}^{n}$ and in the latter case $\{x^{\prime},z^{\prime}\}$ is an edge of $K_{t}^{n}$. This contradicts the fact that $I$ is an independent set. ∎

Next we show that compressions can only decrease the size of the vertex boundary.

Claim 6**.**

For all $i\in[n]$ and $S\subseteq[t]^{n}$, $|\partial c_{i}(S)|\leq|\partial S|$.

Proof.

Fix $\bar{a}:=a_{1},\ldots,a_{i-1},a_{i+1},\ldots,a_{n}\in[t]$. Consider $T=\{(a_{1},\ldots,a_{i-1})\}\times[t]\times\{(a_{i+1},\ldots,a_{n}\}\subset[t]^{n}$.

Note that for every vertex $v\in[t]^{n}$, $\partial\{v\}\cap T$ either has [math] or $t-1$ elements. Thus, $|T\cap\partial S|\in\{0,t-1,t\}$. We claim that $|T\cap\partial c_{i}(S)|\leq|T\cap\partial S|$ for all $T$.

• •

If $|T\cap\partial S|=0$, then there are no edges between $S$ and $T$ and shifting the vertices of $S$ in the $i$th coordinate cannot change that. Thus, $|T\cap\partial c_{i}(S)|=0$.

• •

If $|T\cap\partial S|=t-1$, then the set $\partial T\cap S$ must be constant in the $i$th coordinate. Thus, $c_{i}(\partial T\cap S)=\partial T\cap c_{i}(S)$ is constant in the $i$th coordinate, so $|T\cap\partial c_{i}(S)|=t-1$.

• •

If $|T\cap\partial S|=t$, then trivially $|T\cap\partial c_{i}(S)|\leq t.$

Thus, summing $|T\cap\mathcal{\partial}c_{i}(S))|\leq|T\cap\partial S|$ across all possible $T$, we have that $|\partial c_{i}(S)|\leq|\partial S|$. ∎

Remark 3*.*

The proof crucially uses the fact that $\partial S$ can include elements of $S$. If we instead had defined the vertex boundary to be $\partial S\setminus S$, there is a simple counterexample. Consider $t=3$ and $n=2$ and $S=\{(1,2),(1,3),(2,1),(3,1)\}.$ Then it is not hard to check that $|\partial S|=|\partial c_{1}(S)|=8$, but $|\partial S\setminus S|=4<5=|\partial c_{1}(S)\setminus c_{1}(S)|$.

2.2. Proof of Theorem 1

Define

[TABLE]

First, we show that $\eta(K_{t}^{n})\leq\eta(t)$. In fact, we show a whole family of equality cases.

Claim 7**.**

For all positive integers $n$ and $t$ such that $t\geq 3$, $\eta(K_{t}^{n})\leq\eta(t)$.

Proof.

For all integers $k\in[n]$, consider $S=\{1\}^{k}\times[t]^{n-k}$. Then $\partial S=\{2,\ldots,t\}^{k}\times[t]^{n-k}$. Thus,

[TABLE]

The lower-bound is more difficult, we first need the following inequality, proved in Appendix A.

Claim 8**.**

Let $t\geq 2$ be a positive integer and let $x\geq y\geq 0$ be real numbers, then

[TABLE]

Lemma 9**.**

For positive integers $n\geq 1$ and $t\geq 3$ and all $S\subseteq[t]^{n}$, we have that

[TABLE]

Therefore $\eta(K_{t}^{n})\geq\eta(t)$.

Proof.

By Claim 6 and Remark 2, it suffices to consider the case that $S$ is compressed. We now proceed by induction on $n$.

For our base case, $n=1$, we must have that $S=\emptyset$ in which case (8) is trivial, or $S=[k]$ for some positive integer $k\leq t$. If $S=[1]$, then $\partial S=\{2,\ldots,t\}$, in which case we have an equality case of (8) by the proof of Claim 7. Otherwise, if $k\geq 2$, then $\partial S=[t]$, so $\mu(\partial S)=1$, so (8) holds.

For $n\geq 2$, assume by the induction hypothesis that (8) is true for all $S\subseteq\mathbb{Z}_{t}^{m}$ where $1\leq m<n$. For all $i\in[t]$, let

[TABLE]

Since $S$ is compressed for all $1\leq i\leq j\leq t$, we have that $S_{i}\supseteq S_{j}$. Thus, if $i\in\{2,\ldots,t\}$ is nonzero, for any $x\in(\partial S)_{i}$, there is $y\in S_{0}$ connected to $x$ by an edge of $K_{t}^{n-1}$. Thus, $\partial S_{0}\subseteq(\partial S)_{i}$. Similarly, for any $x\in(\partial S)_{0}$, there is $y\in S_{1}$ such that $x$ is disjoint from $y$. Therefore, $\partial S_{1}\subseteq(\partial S)_{0}$. Putting these together,

[TABLE]

where we applied the inductive hypothesis in the last step. Applying Claim 8, using the fact that $0\leq\mu(S_{1})\leq\mu(S_{0})$, we have that

[TABLE]

as desired. ∎

Claim 7 and Lemma 9 together imply Theorem 1.

2.3. A fine-tuned understanding of the isoperimetric

profile.

Recall the (vertex) isoperimetric profile of a graph $G$ to be

[TABLE]

For $t\geq 3$ fixed, define

[TABLE]

Note that $\Phi_{t}$ is non-decreasing. It is easier to work with $\Phi_{t}(\nu)$ instead of each $\Phi(K_{t}^{n},\nu)$ directly to avoid complications with the discrete behavior of $\Phi(K_{t}^{n},\nu)$ when $n$ is small. By Theorem 1,

[TABLE]

This is tight whenever $\nu=t^{-k}$ for any integer $k\geq 0$, but ceases to be tight when $\log_{t}(\nu)$ is non-integral (see Figure 1).

The following recursive relationship allows one to compute $\Phi_{t}(\nu)$ to arbitrary precision.

Theorem 10**.**

For all $t\geq 3$,

[TABLE]

Using the simple fact that $\Phi_{t}(0)=0$ and $\Phi_{t}(1)=1$, the above equation is extremely powerful. For example,

[TABLE]

which is an exact bound compared to $(\frac{5}{9})^{1/\eta(3)}\approx\frac{7.24}{9}$. This recursion is what allowed the creation of Figure 1.

Theorem 10 is proved in Appendix B. This more refined understanding of $\Phi_{t}$ proves critical in the combinatorial proof of Theorem 4.

3. Independent set stability results

3.1. Black-box result for clique tensor powers

First, we show that if a large independent set $I$ is somewhat close to a maximum-sized independent set $J$, then it is really close to $J$. We fix positive integers $n$ and $t\geq 3$.

Lemma 11**.**

Let $I\subset[t]^{n}$ be an independent set with $\epsilon:=1-t\mu(I).$ Assume there exists a maximum-sized independent set $J$ such that

[TABLE]

Then,

[TABLE]

Proof.

Without loss of generality, we may assume that $J=[t]^{n-1}\times[1]$. Pick $J^{\prime}=[t]^{n-1}\times\{j\}$ such that $j\neq 1$ but otherwise $\mu(I\cap J^{\prime})$ is maximal. Let $\delta:=\mu(I\setminus J)$. Since $J$ and $J^{\prime}$ are disjoint, we have that

[TABLE]

Now, consider $S=\partial(I\cap J^{\prime})$. Recall the definition of $S_{k}\subseteq[t]^{n-1}$ from (9). Since $I\cap J^{\prime}\subseteq J$ has the property that every element has the same last coordinate, $S_{k}=S_{k^{\prime}}$ for all $k,k^{\prime}\neq j$ and $S_{j}=\emptyset$. Thus, $\mu(S_{k})=\frac{t}{t-1}\mu(S)$ for all $k\neq j$. Therefore,

[TABLE]

Applying Theorem 1, we get that

[TABLE]

Since $I$ is an independent set, $\partial I$ is disjoint from $I$. Since $S\cap J=\partial(I\cap J^{\prime})\cap J\subseteq\partial I$, we have that $I\cap J$ and $S\cap J$ are disjoint. Therefore,

[TABLE]

But, we also know that

[TABLE]

By (13) and (14)

[TABLE]

Thus,

[TABLE]

Consider Figure 2 which has a plot of the RHS of (15) when $t=3$. If $\epsilon$ is sufficiently small, then the inequality holds only when $\delta$ is very small (polynomial in $\epsilon$) or very large (about $\frac{1}{t}$). Since is ‘moderately’ small ($\delta\leq\frac{1}{t^{3}}$), we must have that $\delta$ is very small. Quantitatively, note that

[TABLE]

So

[TABLE]

Therefore,

[TABLE]

where the last inequality follows from the following claim which is proved in Appendix A.

Claim 12**.**

For all $t\geq 3$,

[TABLE]

∎

We now use this lemma to ‘amplify’ Theorem 2 to prove Theorem 3.

Proof of Theorem 3..

Set $\epsilon_{t}:=\frac{1}{C_{t}t^{3}}>0$. Consider any independent set $I$ of of $K_{t}^{n}$ such that $\epsilon:=1-t\mu(I)<\epsilon_{t}$. Pick any maximum-sized $J$ guaranteed by Theorem 2 such that

[TABLE]

By Lemma 11, we have that

[TABLE]

as desired. ∎

3.2. Improved stability result for clique tensor powers

In this section we improve $\epsilon_{t}$ in Theorem 3 to an explicit expression. In fact, we may show that

[TABLE]

which corresponds to independent sets $I$ for which $\mu(I)>\frac{3t-2}{t^{3}}$.

First, we try to show that if an independent set $I$ is large enough, then $I$ is either very close to or very far from a maximum-sized independent set. To do this, we show that if $I$ is ‘moderately far’ from a maximum-sized independent set, then this moderate-sized portion which is not in the maximum-sized independent set has such a large vertex boundary that it precludes a large portion of the maximum-sized independent set from being part of $I$, forcing the density of $I$ to be at or below our threshold of $\frac{3t-2}{t^{3}}$.

We need a notation for the maximum sized independent sets. For all $i\in[t]$ and $j\in[n]$ let

[TABLE]

We say that $I$ is sorted if there exists that for all $i_{1},i_{2}\in[t]$ and $j\in[n]$ we have that $i_{1}\leq i_{2}$ implies that

[TABLE]

Note that unlike compressions, we may assume without loss of generality that $I$ is sorted since permuting the labels so that an independent set is sorted does not change its intersection sizes with the maximum independent sets.

Claim 13**.**

Let $I\subset\mathbb{[}t]^{n}$ be a sorted independent set such that $\mu(I)>\frac{3t-2}{t^{3}}$ (or $1-t\mu(I)<\epsilon_{t}$), then for all $j\in[n]$,

[TABLE]

Proof.

Without loss of generality, we may let $j=n$. Denote $J:=J_{1,j}$. Let $\delta=\mu(I\setminus J)$. Since $I$ is an independent set

[TABLE]

Note that $\mu(\partial(I\cap J_{2,n})\cap J_{i,n})$ is [math] if $i=2$ but is $\frac{1}{t-1}\mu(\partial(I\cap J_{2,n}))$ otherwise (see the proof of Theorem 3 for more explanation). Thus, by Theorem 1,

[TABLE]

Since $\mu(I)>\frac{3t-2}{t^{3}}$, we have that

[TABLE]

Thus, we obtain that

[TABLE]

Note that the two sides of the inequality are equal at $\delta_{j}=\frac{t-1}{t^{4}}$ and $\delta_{j}=\frac{t-1}{t^{3}}$. Note that since $1/\eta(t)\in(0,1)$ for all $t\geq 3$, the RHS of (22) is concave for all $\delta\geq 0$. Thus, (22) is false when $\delta\in[\frac{t-1}{t^{4}},\frac{t-1}{t^{3}}]$. Therefore, we have (18). ∎

From Theorem 10, we can attain a bound that is even better.

Claim 14**.**

Let $I\subset[t]^{n}$ be a sorted independent set such that $\mu(I)>\frac{3t-2}{t^{3}}$, then for all $j\in[n]$,

[TABLE]

Proof.

Again, we may assume without loss of generality that $j=n$, let $J=J_{1,j}$. Let $\delta=\mu(I\setminus J)$. From Claim 13, we only need to consider the case that

[TABLE]

From (20)

[TABLE]

Now make the substitution

[TABLE]

where $\delta^{\prime}\in(0,\frac{t-1}{t}]$. From Theorem 10,

[TABLE]

Hence, since $\mu(I)>\frac{3t-2}{t^{3}}$,

[TABLE]

Rearranging,

[TABLE]

Like in the proof of Claim 13, we have equality when $\delta^{\prime}=0$ and $\delta^{\prime}=\frac{t-1}{t}$. Furthermore, since $1/\eta(t)\in(0,1)$ for all $t\geq 3$, the RHS is concave when $\delta^{\prime}\geq 0$. Thus, the inequality is false for all $\delta\in(0,\frac{t-1}{t}].$ Therefore, (24) can never hold, proving (23), as desired. ∎

The next key step is to show Theorem 4 essentially holds for compressed independent sets $I$.

Lemma 15**.**

Let $I\subset\mathbb{[}t]^{n}$ be a compressed independent set such that $\mu(I)>\frac{3t-2}{t^{3}}$, then for some $j\in[n]$,

[TABLE]

Note that by Lemma 11, we immediately have that Theorem 4 holds for compressed independent sets.

Proof.

We prove this statement by induction on $n$. If $n=1$, then the bound holds since $I=\{(1)\}$ which is clearly a maximum-sized independent set. Now assume $n\geq 2$ and that the (25) holds for all compressed independent sets $I\subset\mathbb{[}t]^{n-1}$ with $\mu(I)>\frac{t-1}{t^{3}}$.

Fix a compressed independent set $I\subseteq[t]^{n}$ with $\mu(I)\geq\frac{t-1}{t^{3}}$. From Claim 14, if the lemma is false, then we have that for all $j\in[n]$,

[TABLE]

Since $I$ is compressed, this implies that for all such $j$

[TABLE]

Recall that for all $a\in[t]$, $I_{a}=\{(x_{1},\ldots,x_{n-1})\,:\,(x_{1},\ldots,x_{n-1},a)\in I\}\subseteq[t]^{n-1}.$ We claim that $I_{2}$ is an independent set of $K_{t}^{n-1}$. Note that in general $I_{1}$ is not an independent set of $K_{t}^{n-1}$. Since $I$ is compressed, $I_{2}\subseteq I_{1}$. Thus, if there were $x,y\in I_{2}$ which form an edge of $K_{t}^{n-1}$, then $(x,1),(y,2)\in I$ form an edge of $K_{t}^{n}$, contradicting that $I$ is an independent set. Therefore, $I_{2}\subseteq[t]^{n-1}$ is indeed an independent set.

Note that $\mu(I_{2})=t\mu(I\cap J_{2,n})>\frac{2t-1}{t^{3}}$ which is not sharp enough of a lower bound to invoke the inductive hypothesis. But, we claim that we can find a compressed independent set $\tilde{I}\subseteq I_{1}$ such that $\mu(\tilde{I})\geq\mu(I)>\frac{3t-2}{t^{3}}$.

Pick $a\in[t]$ such that $(I_{1}\setminus I_{2})\cap J_{a,n-1}\subseteq[t]^{n-1}$ has maximal size.222To keep notation as concise as possible, we use the $J_{i,j}$ notation to refer to both the maximal independent sets of $[t]^{n-1}$ and $[t]^{n}$. It should be clear from context which we are referring to. Note that since $I_{1}\setminus I_{2}$ is not necessarily compressed, $a$ might not equal $1$. Let $\hat{I}=I_{2}\cup((I_{1}\setminus I_{2})\cap J_{a,n-1})$. We claim that $I$ is an independent set (although it might not be compressed). As previously established $I_{2}$ is an independent set and clearly $(I_{1}\setminus I_{2})\cap J_{a,n-1}$ is an independent set since the last coordinate is constant. Thus, if $I$ were not an independent set then, there is $x\in I_{2}$ and $y\in I_{1}\setminus I_{2}$ which are connected by an edge in $K_{t}^{n-1}$. But, note that $(x,2),(y,1)\in I$ are connected by an edge in $K_{t}^{n}$, contradiction. Thus, $\hat{I}$ is an independent set of $K^{n-1}_{t}$.

Let $\tilde{I}$ be a compression of $\hat{I}$. since $I_{2}$ and $I_{1}$ are already compressed and $I_{2}\subseteq\hat{I}\subseteq I_{1}$, we have that $I_{2}\subseteq\tilde{I}\subseteq I_{1}$. Now,

[TABLE]

Thus, we may now invoke the induction hypothesis on $\tilde{I}$. Therefore, there exists $j\in[n-1]$ such that

[TABLE]

Since $I_{2}\subseteq\tilde{I}$, we have that

[TABLE]

Therefore, since $I$ is compressed

[TABLE]

Hence, recalling that $I$ is very far from $J_{1,n}$

[TABLE]

Likewise,

[TABLE]

Let $I^{\prime}=I\cap J_{2,j}\cap J_{1,n}$ and $I^{\prime\prime}=I\cap J_{1,j}\cap J_{2,n}$. Now observe that since $I$ is compressed

[TABLE]

Similarly,

[TABLE]

Since $I^{\prime}$ is constant in both the $j$th and $n$th coordinates,

[TABLE]

From Theorem 10, we have that

[TABLE]

since $\Phi_{t}(\nu)\geq 0$. Therefore, since $I^{\prime}\cup I^{\prime\prime}$ is an independent set

[TABLE]

contradiction. Thus, the lemma is true. ∎

Now we extend this result to sorted independent sets; and thus all independent sets.

Lemma 16**.**

Let $I\subset[t]^{n}$ be a sorted independent set such that $\mu(I)>\frac{3t-2}{t^{3}}$, then for some $j\in[n]$,

[TABLE]

Proof.

Like in the proof of Lemma 15, by Claim 14, we may assume for sake of contradiction that for all $j\in[n]$,

[TABLE]

It is not hard to see that for all $i,j\in[n]$ such that $i\neq j$,

[TABLE]

We seek to show that for all $j\in[n]$,

[TABLE]

By Claim 14, assume for sake of contradiction that

[TABLE]

for some $j\in[n]$. We may assume without loss of generality that $j=n$. Since $I$ is sorted,

[TABLE]

Therefore,

[TABLE]

This implies that

[TABLE]

Observe that since $I$ is an independent set

[TABLE]

Therefore, if $x\in\partial(I\cap J_{2,n})\cap c_{n}(I)$, then $(x_{1},\ldots,x_{n-1},1)\in I$ (because any other choice for the last coordinate would violate the above relation). Therefore,

[TABLE]

From this, we get that

[TABLE]

Next, we deduce

[TABLE]

Let $\nu:=\mu(I\cap J_{2,n})$. Then note that

[TABLE]

Thus, by (44)

[TABLE]

We divide the remainder of the proof into three cases depending on the value of $\nu$.

Case333Recall that $\nu>\frac{2t-1}{t^{4}}$ by (37) 1: $\frac{2t-1}{t^{4}}<\nu\leq\frac{1}{t^{2}}$. By Theorem 10 and the fact that $\Phi_{t}(\rho)\geq\rho$ for all $\rho\in[0,1]$,

[TABLE]

Thus, by (45)

[TABLE]

Rearranging,

[TABLE]

This implies that

[TABLE]

Thus, $t^{3}-4t^{2}+8t-2<0$, but this is false for $t\geq 3$, contradiction.

Case 2, $\frac{1}{t^{2}}<\nu\leq\frac{(2t-1)(t-1)}{t^{4}}$.

Then $\Phi_{t}(\nu)\geq\frac{(t-1)^{2}}{t^{2}}$. Thus, by (45)

[TABLE]

contradiction.

Case 3, $\nu>\frac{(2t-1)(t-1)}{t^{4}}$.

Observe that

[TABLE]

Since $I$ is sorted, $\mu(I)\geq 2\nu$. Therefore,

[TABLE]

Thus, $\nu<\frac{4t-2}{t^{4}}$, but $\frac{4t-2}{t^{4}}\leq\frac{(2t-1)(t-1)}{t^{4}}$ for $t\geq 3$, contradiction.

End Cases.

Therefore, our assumption that (35) failed to hold is false. Therefore

[TABLE]

for all $i,j\in[n]$. Applying this fact repeatedly, we can find a compressed $I^{\prime}$ of the same cardinality as $I$ such that $\mu(I^{\prime}\setminus J_{1,i})>\frac{(2t-1)(t-1)}{t^{4}}$ for all $i\in[n]$, contradicting Lemma 15. Thus, our counterexample $I$ could have never existed. This proves the Lemma. ∎

Proof of Theorem 4.

Let $I\subset[t]^{n}$ be an independent set with $\mu(I)>\frac{3t-2}{t^{3}}$. Assume without loss of generality that $I$ is sorted. By Lemma 16, we know that there is $j\in[n]$ such that

[TABLE]

Thus, by Lemma 11, we have that

[TABLE]

as desired. ∎

Acknowledgments

The author is indebted to Venkatesan Guruswami for numerous insightful discussions and comments, in particular for pointing the author to [ADFS04].

The author would also like to thank Boris Bukh and Po-Shen Loh for helpful comments and discussions.

The 2D plots were created using Matplotlib [Hun07]. The 3D visualizations were created using Asymptote [HBP14].

Appendix A Proofs of algebraic inequalities

Proof of Claim 8.

For $c\geq 0$, let $f_{c}(z)=(z+c)^{\alpha(t)}-z^{\alpha(t)}$. Notice that if $z>0$, then $f^{\prime}_{c}(z)=(\alpha(t))((z+c)^{\alpha(t)-1}-z^{\alpha(t)-1})\leq 0$. Thus, we have that $(t-1)f_{c}(y)\geq(t-1)f_{c}(x)$ for all $c\geq 0$. Consider $c=(t-1)y$; we then have that

[TABLE]

Rearranging, we obtain (7). ∎

Proof of Claim 12..

First, verify the cases $t=3$ and $t=4$ using a calculator. Notice that $\eta(t)=\frac{\log t}{\log t-\log(t-1)}\leq t\log t$ so

[TABLE]

Also use a calculator to verify that $h(t):=\frac{t^{2}\log t}{(t-1)^{3}-t}$ is less than $1$ for $t=5$. Now observe that when going from $t$ to $t+1$, the numerator increases by

[TABLE]

and the denominator increases by

[TABLE]

Since $2t^{2}+2t\leq 3t^{2}-3t$ for all $t\geq 5$ and $h(5)\leq 1,$ we have by a simple inductive proof that $h(t)\leq 1$ for all $t\geq 5$. Thus, for all $t\geq 5$,

[TABLE]

as desired. ∎

Appendix B Proof of Theorem 10

The first step in proving this theorem is to determine the structure of $S$ when $\mu(S)$ is fixed but $\mu(\partial S)$ is minimized. In particular, we need $S$ to look as much like a maximal independent set (e.g., $J=[t]^{n-1}\times[1]$) as possible.

Claim 17**.**

Let $t\geq 3$ and $n$ be positive integers. Let $J$ be a maximum-sized independent set. Consider $S\subseteq\mathbb{[}t]^{n}$.

1. (1)

If $\mu(S)<\frac{1}{t}$, then there exists $S^{\prime}\subset[t]^{n}$ such that $\mu(S^{\prime})=\mu(S)$, $\mu(\partial S^{\prime})\leq\mu(\partial S)$, and $S^{\prime}\subset J$. 2. (2)

If $\mu(S)\geq\frac{1}{t}$, then there exists $S^{\prime}\subset[t]^{n}$ such that $\mu(S^{\prime})=\mu(S)$, $\mu(\partial S^{\prime})\leq\mu(\partial S)$, and $J\subseteq S^{\prime}$.

For each $x\in S$, define $|x|$, the level of $x$, be the number of coordinates of $x$ not equal to $1$ (c.f., [ADFS04]).

Proof.

Without loss of generality, assume that $J=[t]^{n-1}\times[1]$. By Claim 6, we may assume that $S$ is compressed. This immediately resolves the case $n=1$, so we may assume $n\geq 2$.

Consider the map $\Pi:\mathbb{[}t]^{n}\to\{0,1\}^{n}$ such that

[TABLE]

Let $f_{S}:=\mathbf{1}_{\Pi(S)}:\{0,1\}^{n}\to\{0,1\}$ be the indicator function of $\Pi(S)$. Since $S$ is compressed, $f_{S}$ is a monotone Boolean function: $f_{S}(x)\leq f_{S}(y)$ whenever $x_{i}\leq y_{i}$ for all $i\in[n]$.

For all $z\in\{0,1\}^{n}$, let $\neg z$ denote the bitwise complement of $z$. Note that for any $x\in\Pi^{-1}(z)$ and $y\in\Pi^{-1}(z)$, $x$ and $y$ are connected by an edge in $K_{t}^{n}$. Therefore, because $S$ is compressed

[TABLE]

and so

[TABLE]

We now describe an algorithm which modifies $S$ into a compressed $S^{\prime}$ such that $\mu(S^{\prime}\cap J)$ is maximized while keeping $\mu(\partial S^{\prime})\leq\mu(\partial S)$ and $\mu(S)\leq\mu(S^{\prime})$. This algorithm consists of two subroutines.

Filling. See Figure 3. Let

[TABLE]

Note that $S\subseteq\operatorname{fill}(S)$ but $\Pi(S)=\Pi(\operatorname{fill}(S))$, so $\mu(\partial(\operatorname{fill}(S)))=\mu(\partial S)$ by (48).

Note that $\operatorname{fill}(S)$ is compressed since $\mathbf{1}_{\Pi(S)}$ is monotone.

Folding.444Note that this Folding operation is considered another form of compression in the literature, although typically used for Kneser graphs. For example see \urlhttps://gilkalai.wordpress.com/2008/10/06/extremal-combinatorics-iv-shifting/ Assume that $S=\operatorname{fill}(S).$ That is, for each $z\in\Pi(S)$, $\Pi^{-1}(z)\subseteq S.$

The operator $\operatorname{fold}_{A}$ is defined for each subset $A\subseteq[n-1]$.

For each $B\subseteq[n]$ let $\sigma_{B}:\mathbb{\{}0,1\}^{n}\to\mathbb{\{}0,1\}^{n}$ be the operator which negates the elements indexed by $B$

[TABLE]

For any $A\subseteq[n-1]$ let

[TABLE]

Then, we define

[TABLE]

Figures 4 and 5 help to visualize this operator.

First, note that in the case $A=\emptyset$, $F_{\emptyset}=0$ since $S$ is compressed. Thus, since $S=\operatorname{fill}(S)=\Pi^{-1}(\Pi(S))$, $\operatorname{fold}_{A}(S)=S$.

For $A\neq\emptyset$, note that since each element of $x\in\Pi(S)$ either stays the same or is replace by $y\in\Pi(\operatorname{fold}_{A})$ such that $|x|\leq|y|$. Thus, since $|\Pi^{-1}(y)|\geq|\Pi^{-1}(x)|$ for all such $x$ and $y$. we have that $\mu(\operatorname{fold}_{i}(S))\geq\mu(S)$. Furthermore, by (48), if we know that $\operatorname{fold}_{A}(S)$ is compressed, then $\mu(\partial\operatorname{fold}_{A}(S))\leq\mu(\partial S)$.

Thus, it suffices to determine when $\operatorname{fold}_{A}(S)$ is compressed. We claim that this is always the case when $\operatorname{fold}_{B}(S)=S$ for all $B\subsetneq A$.

Claim 18**.**

Let $S\subseteq[t]^{n}$ be compressed and $A\subseteq[n-1]$ nonempty. If $S=\operatorname{fill}(S)$ and $\operatorname{fold}_{B}(S)=S$ for all $B\subsetneq A$, then $\operatorname{fold}_{A}(S)$ is compressed and so by the above discussion $\mu(\operatorname{fold}_{i}(S))\geq\mu(S)$ and $\mu(\partial\operatorname{fold}_{A}(S))\leq\mu(\partial S)$.

Proof.

This is equivalent to showing that $\mathbf{1}_{\Pi(\operatorname{fold}_{A}(S))}=\mathbf{1}_{(\Pi(S)\setminus F_{A})\cup\sigma_{A\cup\{n\}}(F_{A})}$ is monotone. Assume for contradiction that there is $x\in\Pi(\operatorname{fold}_{A}(S))$ and $y\in\{0,1\}^{n}\setminus\Pi(\operatorname{fold}_{A}(S))$. such that $y\leq x$.

First consider the case $x\in\sigma_{A\cup\{n\}}(F_{A})$.Thus, $x_{i}=1$ for all $i\in A$ and $x_{n}=0$. Since $y\leq x$, $y_{n}=0$. Let $z=\sigma_{A\cup\{n\}}(x)\in F_{A}\subseteq\Pi(S)$.

If $y_{i}=0$ for some $i\in A$. Then, $y\leq\sigma_{\{i\}}(x).$ Since we assumed $S=\operatorname{fold}_{A\setminus\{i\}}(S)$, we know that $\sigma_{(A\setminus\{i\})\cup\{n\}}(z)=\sigma_{\{i\}}(x)\in\Pi(S)$. Thus, since $S$ is compressed, $y\in\Pi(S)$. But, $y_{n}=0$, so $y\in\Pi(S)\setminus F_{A}\subseteq\operatorname{fold}_{A}(S)$, contradiction.

Otherwise, $x\in\Pi(S)\setminus F_{A}.$ Since $S$ is compressed and $y\leq x$, we have that $y\in\Pi(S)$. Thus, since $y\not\in\Pi(\operatorname{fold}_{A}(S))$, we have that $y\in F_{A}$. Thus $y_{n}=1$, so $x_{n}=1$. Let $z:=\sigma_{A\cup\{n\}}(y)\not\in\Pi(S)$.

Let $B\subseteq A$ be the coordinates $i\in B$ for which $x_{i}=1$. Then, $z^{\prime}:=\sigma_{B\cup\{n\}}(x)$. Since $x\geq y$, it can be checked that $z^{\prime}\geq z$. Since $S$ is compressed and $z\not\in\Pi(S)$, we have that $z^{\prime}\not\in\Pi(S)$. If $B\subsetneq A$, then this contradicts the fact that $z^{\prime}\in\Pi(\operatorname{fold}_{B}(S))=\Pi(S)$. If $B=A$, then this contradicts the fact that $z^{\prime}=\sigma_{A\cup\{n\}}(x)\in\Pi(S)$ because $x\not\in F_{A}$. ∎

Now that we have defined the operators, we finish the proof. First, set $S^{\prime}=\operatorname{flat}(S)$. Now, topologically sort the subsets of $[n-1]$ by inclusion. For each $A\subseteq[n-1]$, in this topological order, apply $\operatorname{fold}_{A}$ to $S^{\prime}$. If it so happens that applying $\operatorname{fold}_{A}$ causes $\operatorname{fold}_{B}(S^{\prime})\neq S^{\prime}$ for some $B$ earlier in the topological order, we go backtrack to the earliest such $B$.

By Claim 18, we know that $S^{\prime}$ is still compressed after each operation. Note that each time $S^{\prime}$ changes, $\mu(S^{\prime}\cap J)$ strictly increases. Thus, after some finite number of applications of these operations, we will have a compressed $S^{\prime}$ such that for all $A\subseteq[n-1]$, $\operatorname{fill}(S^{\prime})=\operatorname{fold}_{A}(S^{\prime})=S^{\prime}$, $\mu(S^{\prime})\geq\mu(S)$, and $\mu(\partial S^{\prime})\leq\mu(\partial S)$.

Furthermore, since $S^{\prime}=\operatorname{fold}_{[n-1]}(S^{\prime})$, we know that either $J\subseteq S^{\prime}$ or $S^{\prime}\subseteq J$. Now, take any $S^{\prime\prime}\subseteq S^{\prime}$ such that $\mu(S^{\prime\prime})=\mu(S)$ while preserving the property that $J\subseteq S^{\prime\prime}$ or $S^{\prime\prime}\subseteq J$. Since $S^{\prime\prime}\subseteq S^{\prime}$, we have that $\mu(\partial S^{\prime\prime})\leq\mu(\partial S^{\prime})\leq\mu(\partial S),$ as desired. ∎

With this claim proven, we may now prove the theorem.

Proof of Theorem 10..

We divide the proof into four parts.

• •

Part 1: If $\nu<\frac{1}{t}$ then $\Phi_{t}(\nu)\leq\frac{t-1}{t}\Phi_{t}(t\nu).$

Consider any $S\subset[t]^{n}$ such that $\mu(S)\geq t\nu$. Let

[TABLE]

be the set where every element of $S$ has a $1$ appended. Note that

[TABLE]

Therefore,

[TABLE]

Thus,

[TABLE]

where $[t]^{*}:=\bigcup_{n\geq 1}[t]^{n}.$

• •

Part 2: If $\nu<\frac{1}{t}$ then $\Phi_{t}(\nu)\geq\frac{t-1}{t}\Phi_{t}(t\nu).$

Consider any $S\subset[t]^{n}$ such that $\mu(S)\geq\nu$. If $n=1$, then $S=\emptyset$, for which it is trivial that $\Phi_{t}(0)=0$. Thus, assume $n\geq 2$.

If $\mu(S)\geq\frac{1}{t}$, then by Theorem 1,

[TABLE]

Thus, we may assume $\mu(S)<\frac{1}{t}$. By Claim 17 there is $S^{\prime}\in[t]^{n}$ such that $\mu(S^{\prime})\geq\nu$, $\mu(\partial S^{\prime})\leq\mu(\partial S)$ and $S^{\prime}\subseteq[t]^{n-1}\times[1]$. Let

[TABLE]

Intuitively, $S^{\prime\prime}$ is $S^{\prime}$ ‘stacked’ $t$ times. Therefore, $\mu(S^{\prime\prime})\geq t\nu$.Then,

[TABLE]

Therefore,

[TABLE]

Thus,

[TABLE]

• •

Part 3: If $\nu\geq\frac{1}{t}$ then $\Phi_{t}(\nu)\leq\frac{t-1}{t}+\frac{1}{t}\Phi_{t}\left(\frac{t\nu-1}{t-1}\right).$

For any $S\subseteq[t]^{n}$ such that $\mu(S)\geq\frac{t\nu-1}{t-1}$, let $S^{\prime}\subseteq[t]^{n+1}$ be

[TABLE]

Then,

[TABLE]

Therefore,

[TABLE]

Hence,

[TABLE]

• •

Part 4: If $\nu\geq\frac{1}{t}$ then $\Phi_{t}(\nu)\geq\frac{t-1}{t}+\frac{1}{t}\Phi_{t}\left(\frac{t\nu-1}{t-1}\right).$

For any $S\subseteq[t]^{n}$ such that $\mu(S)\geq\nu\geq\frac{1}{t}$, by Claim 17 there is $S^{\prime}\in[t]^{n}$ such that $\mu(S)=\mu(S^{\prime})$, $\mu(\partial S^{\prime})\leq\mu(\partial S)$, and $[t]^{n-1}\times[1]\subseteq S^{\prime}$. Pick $j\in\{2,\ldots,t\}$ such that $\mu(S^{\prime}_{j})$ is maximal.555Note that we did not define $S_{j}$ in (9) for $n=1$. In that case, define $S^{\prime}_{j}$ to be $\emptyset$ if $j\not\in S^{\prime}$ and $\{()\}$ if $j\in S^{\prime}$. It is consistent to define $\mu(\emptyset)=0$ and $\mu(\{()\})=1$. Then

[TABLE]

and also

[TABLE]

Thus,

[TABLE]

Therefore,

[TABLE]

∎

Appendix C Optimality of exponent in Theorem 3

In this appendix, we show in (4) of Theorem 3 that the exponent $\eta(t)=\frac{\log t}{\log t-\log(t-1)}$ is optimal and that the constant factor of $2$ is nearly optimal. In other words, the stability result is optimal up to a constant factor.

Lemma 19**.**

For all $t\geq 3$, there exists an infinite sequence of independent sets $\{I_{n}\}_{n\geq 3}$ such that $I_{n}\subset\mathbb{[}t]^{n}$, $\epsilon_{n}=1-t\mu(I_{n})>0$ tends to [math] as $n\to\infty$, and for any $n$ and any maximum-sized independent set $J_{n}$ of $K_{t}^{n}$,

[TABLE]

Proof.

For $n\geq 3$, consider $J_{n}=[1]\times[t]^{n-1}$ and

[TABLE]

See Figure 6 for a visualization.

One may check that $I_{n}$ is an independent set of $K_{t}^{n}$ and $J_{n}$ is a maximum-sized independent set which minimizes $\mu(I_{n}\setminus J_{n})$. Furthermore,

[TABLE]

Thus,

[TABLE]

Notice that since $t^{1/\eta(t)}=\frac{t-1}{t}.$

[TABLE]

Therefore, raising both sides to the $\eta(t)$ power,

[TABLE]

as desired. ∎