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Let $X = C([0; 1])$. For all $f, g \in X$, we define the metric $d$ by $d(f; g) = \sup_x |f(x) - g(x)|$. Show that $S := \{ f\in X : f(0) = 0 \}$ is closed in $(X; d)$. I am trying to show that $X \setminus S$ is open but I don't know where to start showing that.

I wanna add something more, I have not much knowledge about analysis and I am just self taught of it, what I have learnt so far is just some basic topology and open/closed sets.

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Prove that the map $\varphi : X \to \mathbb{R}$ defined by $\varphi(f) = f(0)$ is continuous. ($\varphi$ is even Lipschitz-continuous)

Then notice that $S = \varphi^{-1}(\{0\})$.

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Plus one! Very nice. Now I feel quite stupid. : ) – Rudy the Reindeer Sep 6 '12 at 14:24
I don't quite understand the method. What i have learn't so far is just some basic topology and open close sets.Is that enough to understand it? – Mathematics Sep 6 '12 at 14:28
@Matt Your proof works well too, and puts the emphasis on the fact that the topology of a metric space is completely characterized by converging sequences. My proof can be adapted to the topology of simple convergence though, which is not metrisable. – Ahriman Sep 6 '12 at 14:29
@Mathematics Do you know what is a continuous map between metric spaces ? Do you know that a map $f$ from $X$ to $Y$ is continuous iff the preimage of every closed set in $Y$ is closed in $X$ ? – Ahriman Sep 6 '12 at 14:31
After reading your proof I added the functional-analysis tag to the OP. I like your proof better than mine because it's functional analysis flavour whereas mine is just metric space flavour : ) – Rudy the Reindeer Sep 6 '12 at 14:32

Here's a direct argument: Suppose $f\in X\setminus S$. Let $d=|f(0)|\ne 0$. Then the open ball of radius $d$ around $f$ is disjoint from $S$.

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i have tried to do in a similar way but i don't know how to show it is disjoint from S – Mathematics Sep 6 '12 at 14:36
@Mathematics If $d(f,g)<d$ then in particular $|f(0)-g(0)|<d$, so $g(0)\ne 0$. – Sean Eberhard Sep 6 '12 at 14:38
I don't know if i get it right. Do you try to show that for any $g \in B(f, \epsilon)$ $g$ doesn't in $S$ and hence the set $B(f,\epsilon)$ is in $X-S$ ??? – Mathematics Sep 6 '12 at 14:55
@Mathematics Yes. – Sean Eberhard Sep 6 '12 at 15:15

You can try to show that $X \backslash S$ is open by a contradiction argument. This proof will only assume the very basics of analysis and topology, namely open sets and metrics.

Suppose $X\backslash S$ is not open. That means there is an $f \in X\backslash S$ such that $B_r(f) \cap S \neq \emptyset$ for all $r$. You probably know this, but $B_r(f) = \{g \in C[0,1] : d(f, g) < r\}$ and is called the "open ball of radius $r$". Now, let $f(0) = y \neq 0$. Since $B_r(f) \cap S \neq \emptyset$ for all $r$, we can take a sequence $\{r_n\}$ such that $r_n \to 0$ as $n \to \infty$. This gives rise to a sequence of $g_n \in S$ such that $d(f, g_n) < r_n$, which is equivalent to saying that $\sup_x|f(x) - g_n(x)| < r_n$. Recall that $g_n(0) = 0$, which means that $|f(0) - g_n(0)| < r_n$. This could definitely be the case, but remember that $f(0)$ is a fixed number $y \neq 0$. Since $\{r_n\} \to 0$, we can choose one specific radius $r_N$ such that $r_N < y$. By our assumption, there exists a $g_N \in S$ such that $|f(0) - g_N(0)| < r_N < y = |f(0) - g_N(0)|$. This is a contradiction and so $X\backslash S$ is open.

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I think this one is a nice method but i feel quite confusing at the same time because when i am not looking to the solution, i don't know how to get the contrary. Is the contrary always linked to $r$ for this type of question? – Mathematics Sep 8 '12 at 8:00
I don't know if it is "always" linked but this question offers us a natural link. Try thinking about it like this. We suppose $S$ is not open. That means there is a "point" (really a function) that you cannot draw a ball around so the ball is inside $S$. This is equivalent to saying that every ball we draw is not completely contained in $S$, and must therefore partially be in $X\backslash S$. We draw smaller and smaller balls, and that is how we get the sequence of radiuses going to 0. – Shankara Pailoor Sep 8 '12 at 15:35
I think in this way after reading your solution. What i am surprised is how you linked the inequality |f(0)-g(0)|<$r_N$<y=|f(0)-g(0)| that's why i guess whether it always link to a similar inequality for this type of question. Anyway, thx for your detailed explaination. – Mathematics Sep 9 '12 at 13:16

I'd try to show that $S$ contains all its limit points. To this end let $f$ be a limit point of $S$. Let $f_n$ be a sequence in $S$ converging to $f$ in the sup norm.

Now we show that $f$ is also in $S$:

By assumption, for $\varepsilon > 0$ you have that $\sup_{z \in [0,1]}|f_n(z) - f(z)| < \varepsilon$ for $n$ large enough. In particular, $|f_n(0) - f(0)| = | f(0)| < \varepsilon$ for $n$ large enough. Now let $\varepsilon \to 0$.

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This is just an idea and I have not worked it out in full detail. If you try this and get stuck let me know, then I'll try to work it out in full detail. – Rudy the Reindeer Sep 6 '12 at 14:21
Uniform convergence implies pointwise convergence, so a classical argument justify your method. – Seirios Sep 6 '12 at 14:35
I added a full proof. But I'd try to see what it looks like. Imagine an $\varepsilon$-tube around $f$ in which all $f_n$ are for $n$ large enough. But all the $f_n$ are $0$ at $0$. As you let $\varepsilon$ tend to zero you see that $f$ has got to be zero at zero, too. – Rudy the Reindeer Sep 6 '12 at 17:39
Why go for a contradiction? $|f(0)| = \lim_{n \to \infty}|f_n(0)| = 0$, as $|f(z) - f_n(z)| \leq \lVert f-f_n\rVert = d(f,f_n)\xrightarrow{n\to\infty} 0$ for every $z$ and in particular for $z = 0$ (note that $X$ was used for the space $C[0,1]$, so $x$ isn't a very good choice). Now go and enjoy your risotto... – t.b. Sep 6 '12 at 17:41
The second part of my argument doesn't even need that, just plug in $z = 0$ and get $0 \leq |f(z)| \leq d(f,f_n) \to 0$ which is basically your contradiction without contradiction. Glad to hear about risotto :) – t.b. Sep 6 '12 at 21:14

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