How do we prove that uniform convergence preserves continuity?

Question

Suppose that $(f^{(n)})_{n=1}^{\infty}$ is a sequence of functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and suppose that this sequence converges uniformly to another function $f:X\to Y$. Let $x_{0}$ be a point of $X$. If the functions $f^{(n)}$ are continuous at $x_{0}$ for each $n$, then the limit function $f$ is also continuous at $x_{0}$

My solution (edit)

According to the comments, we have the following solution.

Let us start with the definition of uniform convergence.

For every $\varepsilon/3 > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that $d_{Y}(f^{(n)}(x),f(x)) < \varepsilon/3$ for every $n\geq N$.

Thus, if we fix $n = N$, since each $f^{(n)}$ is continuous at $x_{0}$, we conclude that for every $\varepsilon/3 > 0$ there is a $\delta > 0$ such that for every $x\in X$ the relation holds \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) < \varepsilon/3 \end{align*}

Hence for every $\varepsilon > 0$ there is a $\delta > 0$ such that for every $x\in X$ we have that \begin{align*} d_{Y}(f(x),f(x_{0})) \leq d_{Y}(f(x),f^{(N)}(x)) + d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) + d_{Y}(f^{(N)}(x_{0}),f(x_{0})) < \varepsilon \end{align*}

whenever $d_{X}(x,x_{0}) < \delta$. Therefore $f$ is continuous at $x_{0}$, just as it has been claimed.

Answer 1

$d_Y(f(x),f(x_0)) \leq d_Y(f(x),f_N(x))+d_Y(f_N(x),f_N(x_0))+d_Y(f_N(x_0),f(x_0))$. We can choose $N$ such that the first and the last term are both less than $\epsilon /3$ (by unform convergence). For this particular $N$ we can use continuity of $f_N$ to find $\delta>0$ such that the middle term is less than $\epsilon /3$ whenever $d_X(x,x_0) <\delta$. Now we get $d_Y(f(x),f(x_0)) < \epsilon$ whenever $d_X(x,x_0) <\delta$.

Answer 2

Let $\epsilon >0$, then there exists $N\in \mathbb N $ and $\delta >0$ such that $d(x,x_0)<\delta \Rightarrow d(f_N(x),f_N(x_0))< \epsilon /3$ and $d(f_N(y),f(y))< \epsilon /3$ for any chosen $y$, therefore

$$ \begin{align*} d(f(x),f(x_0))&\leqslant d(f(x),f_N(x))+d(f_N(x),f(x_0))\\ &\leqslant d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))\\ &< \epsilon \end{align*} $$ that is, we had shown that $d(x,x_0)< \delta \Rightarrow d(f(x),f(x_0))< \epsilon $, so $f$ is continuous at $x_0$.