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One of my topology homework questions this week says the following:

Consider the space of bounded functions $B[0, 1]$ on the interval $[0, 1]$ with the $d\infty$ metric. Prove that in the topology induced by the metric $B[0, 1]$ is connected.

I am not sure how to go about proving this, I tried showing that it is path-connected, but since the functions are not necessarily continuous I can't construct a continuous path between them.

The closest answer I've found so far can be found here, but I'm not sure how to apply it to my problem, again since the functions I'm dealing with are not continuous.

Max
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nik
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    You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions. – Lee Mosher Mar 15 '19 at 16:58

1 Answers1

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Let $f, g \in B([0,1])$. By definition, there exists $M, M'$ such that for all $x \in [0,1]$, $$|f(x)| \leq M \quad \text{and} \quad |g(x)| \leq M'$$

For all $t \in [0,1]$, consider the function $f_t : [0,1] \rightarrow \mathbb{R}$ defined by $$f_t : x \mapsto tf(x) + (1-t)g(x)$$

First, $f_t \in B([0,1])$ for all $t \in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $\max(M,M')$.

Moreover, $f_0 = g$ and $f_1 = f$. To see that $t \mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $\varepsilon > 0$. Define $$\eta = \frac{\varepsilon}{M+M' +1}$$

If $|t'-t| \leq \eta$, then $$d_{\infty}(f_t,f_{t'}) = \sup_{x \in [0,1]} |f_t(x) - f_{t'}(x)| = \sup_{x \in [0,1]} | (t-t')f(x) + (t'-t)g(x)|$$ $$ \leq |t-t'| (M+M') \leq \frac{\varepsilon}{M+M' +1} (M+M') \leq \varepsilon$$

This shows that $t \mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.

TheSilverDoe
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