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I was given this set in my lecture notes $$ E= C([0,1],\mathbb{R}) = \{f:[0,1]\longrightarrow \mathbb{R}; \ f \ is \ continuous \} $$ And told to prove/justify that this is an open set. My real analysis/proof writing skills are rusty, but here is my attempt at starting:

We say that E is an open set if for any function $f \in E$, there exists another positive function, $g>0$ such that $(f-g,f+g) \subset E $. I believe this is true because of the properties of functions: We have $$ (1) \ \ \ (f+g)(x)=f(x)+g(x), \ \forall \ x\in[0,1] $$ $$ (2) \ \ \ \forall \ g \in E, \ \exists \ \ -g \ \ such \ \ that \ \ g(x)+(-g(x))=0 $$ With these facts, we have $(f+(-g))(x):=(f-g)(x) \in E$ and $(f+g)(x) \in E$. So we have this interval $[f-g,f+g] \subset E$.

I see that that [f-g,f+g] is closed, but I need the set to be open. I also understand that my approach may be terribly far from correct. If that is so, I apologize. I would like a push in the right direction (assuming there is some correctness to what I wrote). If everything is absolutely wrong, then please let me know. Helpful tips will be appreciated. Thank you in advance.

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    $\begingroup$ Open set in which metric space? $\endgroup$ – zhw. Sep 17 at 1:04
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Let $g_n(x) = 0$ if $0\leq x <1/2$ and $=1/n$ if $1/2 <x\leq 1.$

None of $g_n$ is continuous, so the sequence is in the complement of your set.

In most metrics, the sequence $\{g_n(x)\}$ converges to a continuous function. Since the complement is not closed, your set isn't open.

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