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

semialgebraicset 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

semialgebraicset 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 $\{(x,x^{\alpha\pm \epsilon}):x\ge 0\}\subseteq B$.