Let $(X,d)$ be a compact metric space and let $S= \{f \in C(X):\|f\|\le 1\}$ be the closed unit ball of $C(X)$. Show that if $X$ is an infinite set then $S$ will not be compact.
for infinite compact set $X$ the closed unit ball of $C(X)$ will not be compact
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analysis
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0hmmm it seems to be use full,can you explain a bit more so that i can easily do my work on it? – 2012-10-05
1 Answers
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Let $a$ be an accumulation point of $X$. Let $f_n:X\to\mathbb{R}$ be given by $f_n(x)=\max\big\{1-n~|a-x|,0\big\}.$
You can verify that $\|f_n\|=1$ for all $n$. If the sequence would have a convergent subsequence, it would converge to a discontinuous function.
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0@AtharRaheelAhmad Every subsequence would converge to$a$function that has the value $1$ at $a$ and $0$ everywhere else. That $a$ is an accumulation point guarantees that such a fnction cannot be continuous and that the sequence cannot be eventually constant. – 2012-10-05