Theorem 8.1-3 in Kreyszig's functional analysis

Question

8.1-3 Theorem (Compactness criterion). Let $X$ and $Y$ be normed spaces and $T: X \to Y$ a linear operator. Then $T$ is compact if and only if it maps every bounded sequence $(x_n) $ in $X$ onto a sequence $(Tx_n)$ in $Y$ which has a convergent subsequence.

I want to prove the converse i.e., suppose that every bounded sequence $(x_n)$ in X is mapped onto a sequence $(Tx_n)$ in $Y$, which has a convergent subsequence.

The book's proof for this direction goes along the following lines: Consider any bounded subset $B\subset X$, and let $(Y_n)$ be any sequence in $T(B)$. Then $Y_n = Tx_n$ for some $x_n \in B$, and ($x_n$) is bounded since $B$ is bounded. By assumption, ($Tx_n$) contains a convergent subsequence. Hence, $\overline {T(B)}$ is compact. I don't understand how this compactness follows.

In a normed space $X$, compactness= sequential compactness. So to show that $\overline {T(B)}$ is compact, one should show that every sequence in $\overline {T(B)}$ has a convergent subsequence. But the proof above shows this only for $T(B)$ so I don't understand how the conclusion is made for $\overline {T(B)}$.

Answer 1

The argument seems to be missing one step. To now show that $\overline{T(B)}$ is compact, fix a sequence $(y_n)$ in $\overline{T(B)}$. For each $n\in\mathbb N$, fix some $Y_n\in T(B)$ such that $\|y_n-Y_n\|<\frac{1}{n}$. By the argument in your post, the sequence $(Y_n)$ admits a convergent subsequence, call it $(Y_{n_k})$. Then $(y_{n_k})$ converges to the same limit.

Answer 2

The Author left a little work for the reader implicitly.

What he have proved so far: Given any sequence $(y_n) \subset T(B) $ , $\exists (y_{n_k}) \subset (y_n) $ such that $y_{n_k}\to y\in T(B) $.

What he left for the reader : $(z_n)\subset \overline{T(B) }\implies \exists (y_n^{(k)})\subset T(B) $ such that $y_n^{(k)}\to y_n$