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$X=\{C[0,1],||_{\infty}\}$

$F=\{f\in X : f(1/2)=0\}$

$G=\{g\in X : g(1/2)\neq 0\}$

I need to find which one is open and which one is closed set in $X$.

Well $F$ is closed I guess, as if say $h(x)$ be a limit point of $F$ so there exist sequence $s_n(x)\in F$ such that $s_n(x)\rightarrow h(x)$ so $lim_{n\rightarrow\infty}s_n(x)=h(x)\forall x\in[0,1]$ so $h(1/2)=0$ so $h(x)\in F$ so $F$ is closed and $F^c=G$ is open.

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2 Answers 2

up vote 2 down vote accepted

There is a simpler argument: take $ev_{1/2}$ to be the function $X\to \mathbb R$ defined by $ev_{1/2}(f) = f(1/2)$. Prove that $ev_{1/2}$ is continuous (not hard, and useful in other contexts). Then both $F$ and $G$ are the continuous pre-images of closed and open sets, respectively, hence are closed resp. open.

(If you weren't aware of this general property of continuous functions and open/closed sets, it's pretty useful, worth remembering, and more important if you go on to study general topology).

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$F$ is closed. Suppose $h \in C[0,1]$ such that there exists a sequence $f_n$ converging to $h$ in the supremum norm. You seek to show that $h \in F$. For all $\epsilon > 0$, choose a $N$ such that for all $n > N$, for all $x \in [0,1]$, $|f_n(z) - h(z)| < \epsilon$. Thus $|h(\frac{1}{2}) - f_n(\frac{1}{2})| = |h(\frac{1}{2})| < \epsilon$ for all $\epsilon$. Since $\epsilon$ is arbitrary, $h(\frac{1}{2}) = 0$. Thus $h \in F$.

$G$ is open since it is the complement of $F$ in $C[0,1]$.

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