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Let $(C[0,1],d^*)$ be the metric space defined by: $C[0,1]$ to be a set of all continuous functions from $C[0, 1]$ to $\mathbb{R}$ and $d^*(f, g) = sup\{ |f(x) - g(x)| : x \in [0, 1] \}$

Let $B = \{f : f \in C[0, 1]\ and\ d^*(f, 0) \leq 1\}$ be a closed unit ball.

How can I prove that $B$ is closed?

Similar to the proof for the sequences in $l^{\infty}$, I want to show that every Cauchy sequence of $B$ converges to a point in $B$. Let $\{f_n\}$ be a Cauchy sequence in $B$. Then $\{f_n(x)\}$ is a Cauchy in $\mathbb{R}$ for any $x \in [0, 1]$ - pointwise limit and, since $\mathbb{R}$ is a complete metric space, $f_n(x) \to z_x \in \mathbb{R}$. But then how can I show that set of points $z_x$ forms a continuous function?

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Andreo
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    The completeness part can be found here https://math.stackexchange.com/questions/2164642/proof-of-uniform-limit-of-continuous-functions?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa and the part about being closed can be found here https://math.stackexchange.com/questions/661759/a-closed-ball-in-a-metric-space-is-a-closed-set?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – Severin Schraven May 08 '18 at 21:33

2 Answers2

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If you already know, that $\|.\|_\sup$ norm on $C[0,1]$ is indeed a norm, the proof may be very easy.

It is continuous, because $\bigl|\|x\|−\|y\|\bigr|≤\|x−y\|$. But the unit ball is $\|.\|^{-1}([0,1])$, so the unit ball is closed.

Przemysław Scherwentke
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Hint.-Let $C$ be the complement of $B$ so $g_0\in C\iff d^*(g_0,0)\gt1\Rightarrow d^*(g,0)=1+\epsilon $. Try to show that the (by definition of metric $d^*$) open boule $B_1=\{f\in C[0,1]: d^*(f,g_0)\lt\dfrac{\epsilon}{2}\}$ is such that $B_1\cap B=\emptyset$.

Piquito
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