gpt pro 5.4 attempt on Math-5867

Round-2 continuation for Erdős Problems 741, 749, 750

This is a genuine continuation of the previous round. I only use Round-1 material when explicitly indicated below, and I only re-prove statements when they are strengthened.

Conventions for this continuation: $\overline d(\cdot)$ and $\underline d(\cdot)$ denote upper and lower asymptotic density in $\mathbb N$, and when discussing Problem 750 it is enough to work with induced subgraphs, since deleting edges can only increase the independence number.

Problem 750

1) ROUND-2 OBJECTIVE

Path (C): correct a genuine understatement in Round-1 and then push the known result strictly further.

Round-1 correctly proved the linear-error case using Erdős--Hajnal--Szemerédi (EHS), but it stopped too early. In fact the same EHS input already yields a graph of infinite chromatic number for which every $m$-vertex subgraph has an independent set of size \[ \frac m2-o(m). \] So the original problem is still open, but the currently proved regime is strictly stronger than the Round-1 summary claimed.

2) Round-1 FOUNDATION USED

I use the following Round-1 material.

• Round-1 already explained the linear implication \[ \text{``induced bipartite subgraph on }(1-\delta)m\text{ vertices''} \Longrightarrow \alpha(H)\ge \left(\frac12-\frac\delta2\right)m. \]
• Round-1 used the disjoint-union trick to pass from arbitrarily large finite chromatic number to infinite chromatic number.

I now strengthen that construction by allowing the error parameter to vary from component to component and by choosing the component sizes recursively.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a weighted majorization argument.

For the $j$-th component I choose a local error coefficient \[ \lambda_j:=\frac{1}{(j+1)^2}. \] Then I choose the sizes of the components so large that the cumulative weighted size \[ D_j:=\sum_{i\le j}\lambda_i s_i \] stays below \[ \frac{T_j}{j+1}, \qquad T_j:=\sum_{i\le j}s_i. \] Because the $\lambda_j$ decrease, any $m$-vertex subgraph has maximal possible deficit when it fills the early components first. That turns the componentwise linear estimates into a global $o(m)$ estimate.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was:

Can EHS do anything beyond a fixed proportional error?

Yes. I will prove:

Theorem 750.1. There exists a graph $G$ with $\chi(G)=\infty$ and a function $\eta:\mathbb N\to(0,\infty)$ with $\eta(m)\to 0$ such that every induced subgraph $H$ of $G$ on $m$ vertices satisfies \[ \alpha(H)\ge \frac m2-\eta(m)m. \] Equivalently, there exists a function $g(m)=o(m)$ such that every $m$-vertex subgraph satisfies \[ \alpha(H)\ge \frac m2-g(m). \]

This strictly improves the Round-1 "linear only" conclusion.

5) WORK (ROUND-2)

For each integer $j\ge 1$, set \[ \lambda_j:=\frac{1}{(j+1)^2}. \] By the EHS theorem, for each $j$ there exists a graph $X_j$ with \[ \chi(X_j)>j \] such that every induced subgraph $K\subseteq X_j$ on $t$ vertices contains an induced bipartite subgraph on more than $(1-2\lambda_j)t$ vertices. Because $j$ is finite, the de Bruijn--Erdős theorem implies that $X_j$ has a finite subgraph of chromatic number $>j$; replacing that finite subgraph by the induced subgraph on the same vertex set, we obtain a finite graph $F_j\subseteq X_j$ with \[ \chi(F_j)>j. \] Since the EHS property is hereditary under taking induced subgraphs, every induced subgraph $K\subseteq F_j$ on $t$ vertices still contains an induced bipartite subgraph on more than $(1-2\lambda_j)t$ vertices. Therefore \[ \alpha(K)\ge \left(\frac12-\lambda_j\right)t \qquad\text{for every induced }K\subseteq F_j. \] Let $v_j:=|V(F_j)|$.

For each $j$, let $n_j$ be a positive integer to be chosen recursively, and let $G_j$ be the disjoint union of $n_j$ copies of $F_j$. Then \[ s_j:=|V(G_j)|=n_jv_j, \qquad \chi(G_j)>j, \] and the same local estimate holds on $G_j$: every induced subgraph $K\subseteq G_j$ on $t$ vertices satisfies \[ \alpha(K)\ge \left(\frac12-\lambda_j\right)t. \] Indeed, $K$ splits across the copies, and independence numbers add over disjoint components.

Now define \[ T_j:=\sum_{i=1}^j s_i, \qquad D_j:=\sum_{i=1}^j \lambda_i s_i. \] I claim that we can choose the integers $n_j$ (equivalently, the multiples $s_j$ of $v_j$) so that \[ D_j\le \frac{T_j}{j+1} \qquad\text{for all }j\ge 1. \]

Why this recursive choice is possible. Assume $s_1,\dots,s_{j-1}$ have already been chosen. Then $T_{j-1}$ and $D_{j-1}$ are fixed. We need \[ D_{j-1}+\lambda_js_j\le \frac{T_{j-1}+s_j}{j+1}. \] Since $\lambda_j=1/(j+1)^2$, this is \[ D_{j-1}-\frac{T_{j-1}}{j+1} \le s_j\left(\frac{1}{j+1}-\frac{1}{(j+1)^2}\right) = s_j\frac{j}{(j+1)^2}. \] The coefficient on the right is positive, so the inequality holds for all sufficiently large $s_j$. Because $s_j$ is allowed to be any sufficiently large multiple of $v_j$, we can choose such an $s_j$. Thus the recursion is feasible.

Now let \[ G:=\bigsqcup_{j\ge 1}G_j. \] Since $\chi(G_j)>j$ for every $j$, we have \[ \chi(G)=\sup_j\chi(G_j)=\infty. \]

Let $H$ be any induced subgraph of $G$ on $m$ vertices. Write \[ m_j:=|V(H)\cap V(G_j)|, \qquad \sum_{j\ge 1}m_j=m. \] Applying the local estimate inside each component and summing, \[ \alpha(H)\ge \sum_{j\ge 1}\left(\frac12-\lambda_j\right)m_j = \frac m2-\sum_{j\ge 1}\lambda_j m_j. \] So it remains to bound the weighted deficit \[ \Delta(H):=\sum_{j\ge 1}\lambda_j m_j. \]

Because the sequence $(\lambda_j)$ is decreasing, the maximum possible value of $\Delta(H)$ under the constraints \[ 0\le m_j\le s_j, \qquad \sum_{j\ge 1}m_j=m \] is obtained by filling the early components first.

Exchange argument. Suppose $(m_j)$ maximizes $\Delta(H)$. If there exist indices $i<j$ with $m_i<s_i$ and $m_j>0$, then replacing \[ m_i\mapsto m_i+1,\qquad m_j\mapsto m_j-1 \] increases $\Delta(H)$ by $\lambda_i-\lambda_j>0$, contradiction. Hence in a maximizing configuration all components before some index $k$ are full, all components after $k$ are empty, and the $k$-th component may be partially filled.

Therefore, if \[ T_{k-1}<m\le T_k, \] then \[ \Delta(H)\le D_{k-1}+\lambda_k(m-T_{k-1}). \] Using the recursive bound on $D_{k-1}$, \[ \Delta(H)\le \frac{T_{k-1}}{k}+\frac{m-T_{k-1}}{(k+1)^2}. \] Since $T_{k-1}\le m$ and $(k+1)^{-2}\le k^{-1}$, \[ \Delta(H)\le \frac{m}{k}. \] Thus every induced subgraph $H$ with \[ T_{k-1}<|V(H)|\le T_k \] satisfies \[ \alpha(H)\ge \frac{|V(H)|}{2}-\frac{|V(H)|}{k}. \]

Define $\eta(m)$ by \[ \eta(m):=\frac1k\qquad\text{whenever }T_{k-1}<m\le T_k. \] Since $T_k\to\infty$, we have $k\to\infty$ as $m\to\infty$, so \[ \eta(m)\to 0. \] Therefore \[ \alpha(H)\ge \frac m2-\eta(m)m \] for every $m$-vertex induced subgraph $H$ of $G$. Equivalently, with \[ g(m):=\eta(m)m, \] we have $g(m)=o(m)$ and \[ \alpha(H)\ge \frac m2-g(m). \] This proves Theorem 750.1. $\square$

6) ADVERSARIAL VERIFICATION

• Did I need induced subgraphs only? Yes, and that is enough. For a fixed vertex set, deleting edges can only increase the independence number, so induced subgraphs are the worst case.
• Does disjoint union of many copies preserve the local estimate? Yes. Independence numbers add over disconnected components.
• Could a "scattered" choice of vertices across many components have larger deficit than the greedy one? No. The exchange argument shows any maximizer must fill higher-weight components before lower-weight ones.
• Why does $\eta(m)\to 0$? Because the thresholds $T_k$ are strictly increasing to infinity, so the interval index $k=k(m)$ also tends to infinity.
• What exactly was wrong in Round-1? Not a proof error, but a factual underclaim: the EHS input already gives a sublinear $o(m)$ error term, not merely a fixed linear error.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Problem 750 remains open, but Round-2 proves the strictly stronger known statement \[ \exists\,g(m)=o(m)\ \text{such that}\ \alpha(H)\ge \frac m2-g(m) \] for all $m$-vertex subgraphs $H$ of some graph $G$ with $\chi(G)=\infty$.

So the true unresolved gap is no longer "linear versus sublinear"; it is:

Can one prescribe an arbitrary diverging additive error $f(m)\to\infty$?

That is a much narrower target than the one Round-1 left.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 82%

9) REFERENCES

• N. G. de Bruijn, P. Erdős, A colour problem for infinite graphs and a problem in the theory of relations, Nederl. Akad. Wetensch. Proc. Ser. A 54 $1951$, 371--373.
• P. Erdős, A. Hajnal, E. Szemerédi, On almost bipartite large chromatic graphs, Annals of Discrete Mathematics 12 $1982$, 117--123.
• T. F. Bloom, Erdős Problem #750, erdosproblems.com.