Exercise 8.2.10

A subset

is bounded if there exists some

and some finite bound

such that

. Equivalently, the

can be replaced with

, since given

and a bound

,

and

will also work. Suppose

is not bounded, and let

be some element in

. Then construct the sequence

satisfying

. Clearly no subsequence of

can converge, and therefore

is not compact. Thus, if

is compact then

must be bounded.

Let $k\in X$ be a limit point of $K$, and construct the sequence $(k_n)\in K$ satisfying $d(k_n,k)<1∕n$. Let $(k_n^{\prime })$ be a convergent subsequence of $(k_n)$; given that $d(k_n^{\prime },k)<1∕n$ we must have that $(k_n^{\prime })\to k$, and hence $k\in K$ and $K$ is closed. We showed $Y$ is closed in 8.2.9 (a), and the definition of $Y$ shows it lies within $V_2(0)$ and is hence bounded. Now consider the sequence $f_n$ where

$(f_n)$ does not converge to a function in $C[0,1]$; in particular, it converges pointwise to

It’s clear intuitively that $(f_n)$ cannot have a convergent subsequence. For formality’s sake, let $(g_m)$ be some subsequence of $(f_n)$. For any $m$, let $g_m$ be the $a$’th element of $(f_n)$ (i.e. $g_m=f_a$, and identify $g_p=f_b$ where $b>2a$. Note that $g_m(1∕(2a))=1∕2$ and $g_p(1∕(2a))=1$, hence $d(g_m,g_p)>1∕2$. Since $m$ was arbitrary, $(g_m)$ cannot be a Cauchy sequence and thus cannot be convergent.