Let $\alpha$ be an irrational number, and consider the set $A=\{(x,x^\alpha),x\ge 0\}\subseteq \mathbb{R}^2$, which is the graph of the function $f(x)=x^\alpha$.

I'm trying to prove/disprove the following:

Let $B$ be a semialgebraic set such that $A\subseteq B$, then there exists $\epsilon>0$ such that $\{(x,x^{\alpha\pm \epsilon}):x\ge 0\}\subseteq B$.

This claim seems obviously true to me, but I don't know it to be true for sure. I'd be happy for any proof/counterexample/direction.

EDIT: As shown in the answer below, the claim is false if we take e.g. $\alpha=\log_2 3$ and the set $B=\{(2,3)\}\cup \{(x,y): x\neq 2\}$.

However, what I intend is for $B$ to be "close" to $A$. As suggested below, a rephrase would be:

Let $B$ be a semialgebraic set such that $A\subseteq B$, and for every $u\in A$ there is an open neighborhood of $u$ in $B$, then there exists $\epsilon>0$ such that $\{(x,x^{\alpha\pm \epsilon}):x\ge 0\}\subseteq B$.

  • $\begingroup$ I think that the last $B$ should be corrected to $B_i$. But even this corrected question has negative answer: your can construct your sets $B_i$, so that $B_i\cap\{(x,y)\in\mathbb R^2:x=2\}=\{(2,3)\}$. $\endgroup$ – Taras Banakh Feb 2 '18 at 6:52
  • $\begingroup$ @TarasBanakh - Thanks! I changed as per your suggestion of open sets. Now let's see you ruin the claim :) $\endgroup$ – Shaull Feb 2 '18 at 17:46

The answer to this question is negative:

Jut take $\alpha$ such that $2^\alpha=3$. Taking into account that $2^n\ne 3^m$ for any natural numbers $n,m$, we can prove that the number $\alpha=\log_23$ is irrational.

Now observe that the set $A=\{(x,x^\alpha):x\ge 0\}$ is contained in the closed semialgebraic set $B=\{(x,y)\in\mathbb R^2:x\le 2,\; y\le 3\}\cup\{(x,y)\in\mathbb R^2:x\ge 2,\;y\ge 3\}$.

On the other hand, for any $\varepsilon\ne0$ the number $2^{\alpha+\varepsilon}\ne 3$, so $\{(x,x^{\alpha+\varepsilon}):x\ge 0\}\not\subset B$.

  • $\begingroup$ @Tarash Banakh - Ugh, yes, you're right of course, but that's cheating :) I'll reform the question, since what I want are sets that are "close" to the graph. $\endgroup$ – Shaull Feb 2 '18 at 6:40
  • $\begingroup$ @Shaull Maybe you should add that $B$ is an open semialgebraic neighborhood of $A$? $\endgroup$ – Taras Banakh Feb 2 '18 at 6:42
  • $\begingroup$ @Tarash Banakh - good suggestion. I added a reformulation, but your's is better. I'll revise later today. Thanks! $\endgroup$ – Shaull Feb 2 '18 at 6:46

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