Proving the following set is a $G_\delta$ set $\{x\in \mathbb{R}, \limsup_{m\to\infty}\vert f_m(x)\vert = +\infty\}.$

Question

Is there something wrong with the second sentence of my proof? Did I define the subsequence clearly? Thank you.

$\def\R\{\mathbb{R}}$

Let $f_m \colon \R\to \R (m=1,2,\dots)$ be a sequence of continuous functions. Prove the following set is a $G_\delta$ set $$\left\{x\in \R, \limsup_{m\to\infty}\vert f_m(x)\vert = +\infty\right\}.$$

$\textit{Proof.}$ Observe that $\limsup_{m \to \infty} |f_m(x)| = +\infty$ when there is a subsequence $n_k$ such that $|f_{n_k}(x)| \to +\infty$ as $k$ tends to $\infty.$

Thus for any boundary $n \in \Bbb N$ we are able to indicate $m > n$ such that $|f_{m}(x)| > m,$ which is equivalent to the subsequence condition.

So $x \in \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty O_m$ where $O_m= \{x: |f_m(x)| > m\}$ is open by continuity of $f_m$ and $|f_m|$. Thus the union, for each $n,$ is open and we see $x$ is equal to an intersection of open sets, hence it is a $G_\delta$.

Answer 1

Yes, that is wrong. Taking $f_m(x)=\frac m 2$ for all $x$ we get a counter-example to your claim.

There is no reason why you should have $|f_m(x)| >m$. Correct proof: The given set is $\bigcap_N \bigcap_k\bigcup_{j\geq k} \{x: |f_j(x)| >N\}$ which is a countable intersection of open sets.