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I'm supposed to find out whether the set

$$F := \{f \in C([a,b]) \colon f(t) > 0 \, \forall t \in[a,b]\}$$

is open in the topology $\mathcal{O}_{d_{\infty}}$, defined by the metric $d_{\infty}(f,g) := \sup\{|f(t)-g(t)|, \, t\in[a,b]\}$. Now my ansatz was to take any function $f\in F$ and check if it has a neighborhood that lies entirely in $F$. But how would I do that?

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  • $\begingroup$ What does a basis neighbourhood of $f$ look like? $\endgroup$ Jan 22 '17 at 15:51
  • $\begingroup$ Something like a "tube" around the values of $f$? $\endgroup$ Jan 22 '17 at 15:55
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If $f$ is continuous and $\forall t \in [a,b]$, $f(t) > 0$, then there exists $\epsilon > 0$ such that $f(t) \ge \epsilon$ for all $t\in [a,b]$. Indeed, if for all $\epsilon > 0$ there exists $t \in [a,b]$ such that $f(t) < \epsilon$, then there exists a sequence $(t_n) \subset [a,b]$ such that $f(t_n) \to 0$. By Bolzano-Weierestrass, $(t_n)$ has a convergent subsequence $(t_{n_k})$ converging to some $t \in [a,b]$, and so $\lim_{x\to t} f(x) = 0$, hence $f(t) = 0$, a contradiction.

Thus, the ball of center $f$ and radius $\epsilon/2$ is contained in $F$.

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