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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: X\to Y$ with $n\in\mathbb{N}$ and $f:X\to Y$ functions. $f_n$ is continuous for every $n$ and $f_n\stackrel{n\to\infty}{\longrightarrow} f$ uniformly.

Then $f$ is continuous.

Here is my proof:

Let $x\in X$ and $\epsilon >0$ be arbitrary. Since $f_n\to f$ uniformly it exists $N\in\mathbb{N}$ such that $d_Y(f_N(x), f(x))<\epsilon/3$ for every $x\in X$.

As $f_N$ is continuous we have for $x_0\in X$ and $\delta > 0$, that $d_Y(f_N(x),f_N(x_0))<\epsilon/3$ if $d_X(x,x_0)<\delta$.

This gives us:

$$\begin{align}d_Y(f(x),f(x_0)&\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\\ &\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\\ &< \epsilon/3+\epsilon/3+\epsilon/3=\epsilon\end{align}$$

Thanks in advance for your correction.

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    $\begingroup$ I see no need to correct whatever (other than a typo). $\endgroup$ Jul 17, 2018 at 18:41
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    $\begingroup$ Thanks for the confirmation. Better safe then sorry. $\endgroup$
    – Cornman
    Jul 17, 2018 at 18:42

2 Answers 2

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The idea of this proof is entirely correct, and it's very well executed.

There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $\delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<\epsilon/3$.

So I would change it to something like this:

Let $x\in X$ and $\epsilon >0$ be arbitrary. Since $f_n\to f$ uniformly, it exists $N\in \Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<\epsilon/3$ for all $x_1\in X$.

As $f_N$ is continuous, there is a $\delta>0$ such that for any $x_0$ with $d_X(x, x_0)<\delta$ we have $d_Y(f_N(x), f(x_0))<\epsilon/3$.

Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $\leq$.

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  • $\begingroup$ Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<\epsilon/3$ holds for every $x\in X$. Espacially for $x_0$. This is given by the uniform convergenc. $\endgroup$
    – Cornman
    Jul 17, 2018 at 18:55
  • $\begingroup$ @Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice. $\endgroup$
    – Arthur
    Jul 17, 2018 at 18:57
  • $\begingroup$ Ok, I will keep that in mind. :) $\endgroup$
    – Cornman
    Jul 17, 2018 at 18:58
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You are close, but in my opinion in one detail you are not sufficiently precise enough.

The problem lies in the inequality $$ d_Y(f_N(x_0),f(x_0)) \leq \varepsilon/3, $$ which you only showed for $x$, not for $x_0$.

In my opinion you should replace the first line of the proof with something like

Let $\varepsilon>0$ be arbitrary. Since $f_n\to f$ uniformly there exists $N\in\mathbb N$ such that $d_Y(f_N(z),f(z))<\varepsilon/3$ for all $z\in X$.

If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.

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  • $\begingroup$ Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $x\in X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from. $\endgroup$
    – Cornman
    Jul 17, 2018 at 18:54

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