appreciate your help with the below:


Let C[0,1] be the set of continuous functions from [0,1] to $\mathbb{R}$. Consider the metric space M = (C[0,1],d) where d denotes the sup metric. Show the set of functions in C[0,1] whose image is contained in (0,1) is an open set in C[0,1].


My attempt:

So we let K denote the set of continuous functions from [0,1] to $\mathbb{R}$ whos image is contained in (0,1). Take an arbitrary function k in K. Then since k is bounded, it attains its bounds. So a and b exists in [0,1] such that 0 < k(a) $\leq$ k(x) $\leq$ k(b) < 1 for x in [0,1]. Pick $\epsilon$ = min{ k(a) , 1-k(b) }, so we have the open ball $B_{\epsilon}(h)$ of radius $\epsilon$ around k.

I am unsure from here onwards. Can you please suggest how should I proceed please?

Many thanks

  • 1
    $\begingroup$ It looks like you just need to push inequalities around to show that $B_\epsilon(k) \subset K$. $\endgroup$
    – Rolf Hoyer
    Apr 4, 2015 at 8:57
  • $\begingroup$ Hi Rolf, thanks for this but can you elaborate a little more please? $\endgroup$
    – beginner
    Apr 5, 2015 at 23:43

1 Answer 1


In order to show that $K$ is open, you have chosen a point $k\in K$, and need to show that $B_\epsilon(k) \subset K$ for some $\epsilon >0$. You have chosen $\epsilon = \min(k(x), 1-k(x))$, which guarantees the inequalities $\epsilon \le k(x)$ and $\epsilon \le 1 - k(x)$ for all $x\in [0,1]$. Note that these inequalities can be rewritten as $0 \le k(x) - \epsilon$ and $k(x) + \epsilon \le 1$. So far, so good.

In order to complete the proof, you must take an arbitrary $g\in B_{\epsilon}(k)$, and show that $g(x) \in (0,1)$ for all $x\in [0,1]$. This would show that $g \in K$, which would then imply $B_\epsilon(k) \subset K$, as desired. I will leave it at that, with a hint to examine the ramifications of the statement '$g\in B_\epsilon(k)$'.


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