Let $\{x_n(t)\}_{n=1}^{\infty}$ be real a sequence of continuous function from $[0,1]$ to $\mathbb{R}$, and $\{x_n(t)\}_{n=1}^{\infty}$ converges pointwise to $x(t)$ i.e. $\lim_{n \to \infty} x_n(t) =x(t)$.
Define $P_{m,n}=\{t||x_m(t)-x(t)\leq 1/n\}$, $Q_n=\bigcup_{m=1}^{\infty} (P_{m,n})^{\circ}$, $R=\bigcap_{n=1}^{\infty}Q_n$.
{$(A)^{\circ}$ is interior of A}
Question:
Show that $R$ consists of all $t$ such that $x(t)$ is continuous.
my thought:
$s \in R \Rightarrow s \in \bigcap_{n=1}^{\infty} \bigcup_{m=1}^{\infty} (P_{m,n})^{\circ}$
I thought $s$ could be written by a form like $(\epsilon, \delta)$ definition of limit, but I had no idea what to do next. I would be really thankful if someone can help me.
$$ s \in R \Leftrightarrow \quad\forall n, \exists m, \delta = \delta(m,n) > 0 \text{ such that } |t-s| < \delta \Rightarrow |x_m(t)-x(t)| < 1/n $$ So $s \in R$, and $\epsilon > 0$, choose $n\in \mathbb{N}$ such that $1/n<\epsilon$, then choose $m\in \mathbb{N}, \delta > 0$ as above. Then, if $|t - s| < \delta$, then $$ |x_m(t) - x(t)| < \epsilon \text{ and } |x_m(s) - x(s)| < \epsilon $$ Now choose $\delta_0 > 0$ using the continuity if $x_m$ and the triangle inequality to get $$ |x(s) - x(t)| < 3\epsilon \text{ if } |t-s| < \min\{\delta,\delta_0\} $$ This proves that $x$ is continuous at $s$.
Now try reversing this argument (using the fact that for any $\epsilon > 0, \exists n\in \mathbb{N}$ such that $1/n<\epsilon$) and let me know if you can prove the converse.