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Consider $X = C[0, 1]$, the space of continuous functions on $[0, 1]$, equipped with the metric $d_\infty(f, g) = \max_{x∈[0,1]} |f(x) − g(x)|$.

Let set $A\subset X$ be defined as $A = \lbrace f : 0 ≤ f(x) < 1 $ for all $ x ∈ [0, 1]\rbrace$.

Use definition to find the closure of $A$.

That is, to find the closure, we should explicitly find the limit points of $A$ with justification.

I have found the functions $A' = \lbrace f : 0 ≤ f(x) \le 1 $ for all $ x ∈ [0, 1]\rbrace$ is the set of limit points of $A$.

Now I have difficulty to prove that the functions with function value $0$ or $1$ in somewhere are limit points of $A$. To explicitly show that, I need to prove that $\forall \epsilon>0, \exists f_0\in B_\epsilon(f)$ such that $f_0\ne f$.

So could someone please tell me how to do that? Thanks!

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  • $\begingroup$ @HennoBrandsma Sorry. It is just an accident... I did not notice that. $\endgroup$
    – Y.X.
    Apr 8 '17 at 11:21
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If $t \in [0,1)$, $d_\infty(f, tf) = (1-t)||f||_\infty$, where $||f||_\infty = \sup_{x \in [0,1]} |f(x)|$, is the norm of $f$. And if $f$ maps into $[0,1]$, $tf$ maps into $[0,t]$,so $tf \in A$ and $(1-\frac{1}{n})f \rightarrow f \text{ as } n \rightarrow \infty)$.

You will need $A^\circ$ too. This will be $B=\{f \in C([0,1]: \forall x: 0< f(x) < 1\}$: $f \in B$ implies $\exists r >0: \forall x: |f(x)| \le 1-r \land |f(x)| \ge r$ (Use that $f$ has a min and max in $(0,1)$), and then $B_{d_{\infty}}(f, \frac{r}{2}) \subseteq B$, showing $B$ is open.

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You need some sort of a bump function. Let $f\in C[0,1]$ satisfy $0\leq f(x)\leq1$ for every $x$. One can show that there is a function $\varphi\in C[0,1]$ satisfying the following conditions:

1) $\max |\varphi(x)|<\epsilon$.

2) For every $x$ such that $f(x)=1$, we have $\varphi(x)>0$.

3) For every $x$ such that $f(x)<\epsilon$, we have $\varphi(x)=0$.

Then, the function $f-\varphi$ gives you what you need.

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  • $\begingroup$ So are we considering the ball centered at $f$ defined as $B_\varphi (f)$? Is the radius of an open ball allowed to be a function here in this space? $\endgroup$
    – Y.X.
    Apr 8 '17 at 1:22
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    $\begingroup$ @PropositionX No, the radius of a ball in a metric space is always a positive number. In this case, the function $f-\varphi$ is contained in the ball $B_\epsilon (f)$. $\endgroup$ Apr 8 '17 at 1:51

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