$f$ is monotone increasing in $\mathbb{R}$ . Prove that the set $\{x: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon) \}$ is closed. $f$ is monotone increasing in $\mathbb{R}$ . Prove that the set $\{x: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon) \}$ is closed.
I want to change the set into $$\cap_{\epsilon\in \mathbb{R}^+} \{x:f(x+\epsilon)\gt f(x-\epsilon)\}$$
and see if there is a correlation between it and the close set. But it's not a intersection of countable interval.
 A: Hint: show instead that the set $$\left\{x\in\mathbb R\,\big{|}\,\exists\,\varepsilon>0:f(x+\varepsilon)\leq f(x-\varepsilon)\right\}$$ is open.
A: Let $(x_n)_n$ be a sequence in $\{x \in \mathbb{R}: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon) \}$ which converges to $x_0 \in \mathbb{R}$. 
We wish to show that $x_0 \in \{x \in \mathbb{R}: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon) \}$.
Let $\varepsilon >0$. Pick $n \in \mathbb{N}$ such that $|x_0 - x_n| < \frac\varepsilon2$. This is equivalent to $x_0 \in \left\langle x_n - \frac\varepsilon2, x_n + \frac\varepsilon2\right\rangle$.
We then have $x_0 + \varepsilon > x_n + \frac\varepsilon2$ and $x_n - \frac\varepsilon2 > x_0 - \varepsilon$. Therefore
$$f(x_0+\varepsilon) \ge f\left(x_n + \frac\varepsilon2\right) > f\left(x_n - \frac\varepsilon2\right) \ge f(x_0-\varepsilon)$$
Since $\varepsilon$ was arbitrary, we conclude $x_0 \in \{x \in \mathbb{R}: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon)\}$.
Therefore $\{x \in \mathbb{R}: \forall \epsilon > 0, f(x+\epsilon)> f(x-\epsilon) \}$ is closed.
A: Hint: Take the complement, and show it contains an interval around each of its points.

 $x$ being in the complement means there is $\varepsilon>0$ with $f(x+\varepsilon)=f(x-\varepsilon)$. Now $f$ must be constant on $[x-\varepsilon, \, x+\varepsilon]$, and thus every point of $(x-\varepsilon, \, x+\varepsilon)$ is in the complement. 

