Forcing $3$-term arithmetic sequences into sets using $\{1,2,3,4 \}$.

Question

Here is the full problem, which I couldn't fit in the title:

A set $D \in \mathbb{Z}^+$ is 'excellent' if it has the following condition: $\forall A \subset \mathbb{Z}$, there exists $a \in \mathbb{Z}$ and $d \in D$ s.t. $\{a-d,a,a+d \} \subset A $ or $\subset \mathbb{Z} \setminus A$. Prove that the set $D= \{1,2,3,4 \}$ is excellent.

This question is a dubious one. Right off the bat, one can observe two cases:

• Case 1: $a \in A$.

• Case 2: $a \in \mathbb{Z} \setminus A$.

It might be beneficial to consider multiple approaches. These could include contradictions or possibly checking sets $D \in \{ \{1\}, \{1,2\}, \{1,2,3\} \}$ to see why they do or don't work.

Note that if $A$ is bounded from above or below, we can simply pick $a = \max \{A \} \pm 2$, so that $\{a-1,a,a+1 \} \subset \mathbb{Z} \setminus A$, and the condition is satisified. This means that for any possible refutation to $D$ being excellent must include an $A$ that extends infinitely in both directions.

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Approach:

Suppose for the sake of argument $A$ is not bounded from below or above. Choose 'center' $a \in A$ such that $a$ does not satisfy the condition as stated in the problem. Therefore: there exists some for no $d \in \{1,2,3,4 \} $ is it true that $a-d, a+d \in A$. (Both simultaneously)

This means that either one is in $A$ and the other in $\mathbb{Z} \setminus A$ or both are $\in \mathbb{Z} \setminus A$. What is particularly interesting is that we can then pick our center as $a-d$ or $a+d$, and essentially repeat the process to claim that there are an infinite elements in both $A$ and $\mathbb{Z} \setminus A$.

Allow $a+d \in A$ ($a-d$ can be used for this claim, also). Then, we can say that either $a$ or $a+2d \in A$, but not both, as that would violate our claim. Because we already know that $a \in A$, we know that $a+2d \in \mathbb{Z} \setminus A$.

Because $d$ is arbitrary in $D$, it is safe to assume that $a+2d \in \mathbb{Z} \setminus A \text{,} \ \ \forall d \in D $. However, this means that $\{a+2, \ a+4, \ a+6 \} \subset \mathbb{Z} \setminus A$, which forces a $3$-term arithmetic progression into $\mathbb{z} \setminus A$ with center $a+4$ and $d=2$. $\blacksquare$ (?)

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Observation

I feel that it's important to notice that we assume $a \in A$, so we can conclude that, because of the unboundedness of $A$, that there are an infinite number of arithmetic progressions in $\mathbb{Z} \setminus A$. Are there any interesting properties that arise from this?

Another observation is that we assumed from the beginning that $a \in A$ did not satisfy the conditions, which there must exist for there to be no valid center in $A$.

My Main Question

I made this post not only to ask you this question, but to share this problem with you. That being said, my question pertains to the switching of centers. Does switching centers invalidate the proof?

Are there any other illogical leaps or refutations to my proof? Help is greatly appreciated.

Answer 1

This problem can be solved by brute force. Since $A$ and $\mathbb{Z} \backslash A$ are symmetric in the problem, we may assume wlog that $0 \in A$.

For a given set $A$, consider the sequence $a_0, a_1, a_2, \cdots$ where $a_i = 1$ if $i \in A$ and $a_i = 0$ if $i \not\in A$. If $A$ is a set that violates the stated condition, then for any $k$ we have that $a_{k - d}, a_k, a_{k + d} \; (d \in \{1,2,3,4\})$ cannot be all 1 or all 0. Thus we can start from the finite sequence $[a_0]$, and attempt to extend the sequence such that the above property is preserved. If all such attempts fail within finite steps, we know that $\{1,2,3,4\}$ is an "excellent" set.

Here's a Python script that does this search:

curr_lists = [[1]];
iter = 0;
while len(curr_lists) > 0:
    iter = iter + 1;
    l = curr_lists[0];
    print(str(l));
    del curr_lists[0];
    must_be_zero = False;
    must_be_one = False;
    k = len(l);
    if k >= 2 and l[k - 1] == 0 and l[k - 2] == 0:
        must_be_one = True;
    if k >= 2 and l[k - 1] == 1 and l[k - 2] == 1:
        must_be_zero = True;
    if k >= 4 and l[k - 2] == 0 and l[k - 4] == 0:
        must_be_one = True;
    if k >= 4 and l[k - 2] == 1 and l[k - 4] == 1:
        must_be_zero = True;
    if k >= 6 and l[k - 3] == 0 and l[k - 6] == 0:
        must_be_one = True;
    if k >= 6 and l[k - 3] == 1 and l[k - 6] == 1:
        must_be_zero = True;
    if k >= 8 and l[k - 4] == 0 and l[k - 8] == 0:
        must_be_one = True;
    if k >= 8 and l[k - 4] == 1 and l[k - 8] == 1:
        must_be_zero = True;
    if must_be_zero == False:
        l_copy = l.copy();
        l_copy.append(1);
        curr_lists.append(l_copy);
    if must_be_one == False:
        l_copy = l.copy();
        l_copy.append(0);
        curr_lists.append(l_copy);
print("Search complete after " + str(iter) + " iterations")

This program terminates after 39 iterations. Therefore $\{1,2,3,4\}$ is an "excellent" set.

Answer 2

First let me restate the problem in a more civilized form:

Show that, in any $2$-coloring of the integers, there is a monochromatic $3$-term arithmetic progression with common difference at most $4$.

The Van der Waerden number $W(2,3)=9$ means that in any $2$-coloring of the numbers $\{1,2,3,4,5,6,7,8,9\}$ there is a monochromatic $3$-term arithmetic progression, which of course has common difference at most $4$.

I don't know any way to prove $W(2,3)=9$ other than by brute force. Fortunately the brute force method is a simple backtrack algorithm which (in this small case) can be carried out by hand in a few minutes. Writing it down in the form of a proof on an examination paper might be a challenge. Maybe there is an easier way to prove the weaker result that the set $\{1,2,3,4\}$ is "excellent".