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I have a doubt regarding the result that uniform convergence preserves continuity.

Statement: If a sequence of continuous functions $\{f_n\}_n$ $f_n : A → \mathbb{R}$ converges uniformly on $A \subset \mathbb{R}$ to $f : A \rightarrow \mathbb{R}$, then $f$ is continuous on $A$.

Question: Is $A\subset \mathbb{R}$ a necessary condition for the statement? Can we have $A=\mathbb{R}$? Why?

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    $\begingroup$ $A \subset \mathbb{R}$ is for most authors the same as $A \subseteq \mathbb{R}$. And besides, the proof works if $A$ is an arbitrary topological space. $\endgroup$ – Daniel Fischer Feb 26 '16 at 14:21
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The standard proof (involving $\epsilon$-s and $\delta$-s) works even if $A=\mathbb R$. The most likely explanation is that $\subset$ in whatever book you are reading actually means $\subseteq$.


But let's assume we have only a proof of the following statement:

If $f_n: A\to\mathbb R$ converges to $f$ uniformly and $A\neq \mathbb R$ and $A\subset \mathbb R$, then $f$ is continuous.

We want to prove the statement:

If $f_n: \mathbb R\to\mathbb R$ converges to $f$ uniformly, then $f$ is continuous.

Proof:

Let's take a sequence of functions $f_n$ such that $f_n:\mathbb R\to\mathbb R$ and $f_n$ converges to $f$ uniformly on $\mathbb R$.

Then, we can define $g_n:[-1,\infty)\to \mathbb R$ as $g_n(x)=f_n(x)$ for every $x\geq 0$. It is easy to show that $g_n$ converges to $g$, defined as $$g:[-1,\infty)\to \mathbb R\\ g(x)=f(x)$$ and, since $[0,\infty)\neq \mathbb R$, we conclude that $g$ is continuous.

Similarly, we can show that $h:(-\infty, 1]\to\mathbb R$, defined as $h(x)=f(x)$, is continuous.


Conclusion: $f$ is continuous.

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Yes $A$ can be the hole $\Bbb R$. Why? Because the proof is valid for $A=\Bbb R$.

For all $x_0\in\Bbb R$ we have

$$\vert f(x)-f(x_0)\vert\le \vert f_n(x_0)-f(x_0)\vert+\vert f_n(x_0)-f_n(x)\vert+\vert f_n(x)-f(x)\vert$$ so by the continuity of $f_n$ and the uniform convergence you can make $\vert f(x)-f(x_0)\vert$ less than any $\epsilon>0$ for a convenient choice of $\delta$ such that $|x-x_0|<\delta$ and a convenient choice of $N$ and let $n\ge N$.

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