[proofplan] We prove both implications from the definition of compactness for operators. If $T$ maps bounded sets to relatively compact sets, then the image of any bounded sequence lies in a compact [metric space](/page/Metric%20Space), and compactness forces a convergent subsequence. Conversely, assuming the subsequence property, we show that the image of every bounded set is totally bounded; since its closure is a closed subset of the complete space $Y$, it is complete, hence compact. [/proofplan]

[step:Extract a convergent subsequence from compactness of the image closure]Assume that $T:X\to Y$ is compact. Let $(x_k)_{k\in\mathbb N}$ be a bounded sequence in $X$, and define the bounded subset \begin{align*} A:=\{x_k:k\in\mathbb N\}\subset X. \end{align*} By compactness of $T$, the set \begin{align*} K:=\overline{T(A)}^{\,Y}\subset Y \end{align*} is compact. Since $Tx_k\in K$ for every $k\in\mathbb N$, it remains only to use the [sequential compactness](/page/Sequential%20Compactness) of compact metric spaces. We verify this directly. Let $(y_k)_{k\in\mathbb N}$ be any sequence in a compact metric space $(K,d)$. For $y\in K$ and $r>0$, define the open ball in $K$ by \begin{align*} B_K(y,r):=\{z\in K:d(z,y)<r\}. \end{align*} If some value $y\in K$ occurs infinitely many times among the $y_k$, then those repeated terms form a subsequence converging to $y$. Otherwise, suppose no subsequence converges in $K$. Then for each $y\in K$ there is a radius $r_y>0$ such that $B_K(y,r_y)$ contains only finitely many terms of the sequence; otherwise one could choose indices $k_j$ strictly increasing with \begin{align*} y_{k_j}\in B_K\left(y,\frac{1}{j}\right), \end{align*} giving $y_{k_j}\to y$. The family \begin{align*} \{B_K(y,r_y):y\in K\} \end{align*} is an open cover of $K$. Compactness gives finitely many points $y_1,\dots,y_m\in K$ such that \begin{align*} K=\bigcup_{i=1}^m B_K(y_i,r_{y_i}). \end{align*} Each ball in this finite union contains only finitely many terms of the sequence, so their union contains only finitely many terms, contradicting that every $y_k$ lies in $K$. Hence every sequence in $K$ has a convergent subsequence. Applying this to $y_k:=Tx_k$ gives a subsequence $(Tx_{k_j})_{j\in\mathbb N}$ converging in $Y$.[/step]

[guided]Assume that $T:X\to Y$ is compact in the operator sense: whenever $A\subset X$ is bounded, the closure of $T(A)$ in $Y$ is compact. Start with an arbitrary bounded sequence $(x_k)_{k\in\mathbb N}$ in $X$. The set of its terms \begin{align*} A:=\{x_k:k\in\mathbb N\}\subset X \end{align*} is bounded because the sequence is bounded. Therefore compactness of $T$ gives compactness of \begin{align*} K:=\overline{T(A)}^{\,Y}\subset Y. \end{align*} Every term $Tx_k$ lies in $T(A)$ and hence in $K$. We now prove the metric fact needed here: every sequence in a compact metric space has a convergent subsequence. Let $(K,d)$ be compact and let $(y_k)_{k\in\mathbb N}$ be a sequence in $K$. For $y\in K$ and $r>0$, define \begin{align*} B_K(y,r):=\{z\in K:d(z,y)<r\} \end{align*} to be the open ball in $K$ with centre $y$ and radius $r$. If some point $y\in K$ occurs as $y_k$ for infinitely many indices, then selecting those indices gives a constant subsequence, which converges to $y$. So assume no value occurs infinitely often. Suppose, toward a contradiction, that $(y_k)$ has no convergent subsequence. For a point $y\in K$, if every ball $B_K(y,r)$ contained infinitely many terms of the sequence, then choosing successively \begin{align*} y_{k_j}\in B_K(y,1/j), \qquad k_1<k_2<\cdots, \end{align*} would give $d(y_{k_j},y)<1/j$, hence $y_{k_j}\to y$. Since no convergent subsequence exists, for each $y\in K$ there must be a radius $r_y>0$ such that $B_K(y,r_y)$ contains only finitely many terms of the sequence. The balls $B_K(y,r_y)$ cover $K$. By compactness, finitely many suffice: \begin{align*} K=\bigcup_{i=1}^m B_K(y_i,r_{y_i}) \end{align*} for some $y_1,\dots,y_m\in K$. Each ball in this finite union contains only finitely many terms of $(y_k)$, so the whole [compact space](/page/Compact%20Space) $K$ contains only finitely many terms of the sequence. This contradicts $y_k\in K$ for every $k\in\mathbb N$. Thus a convergent subsequence must exist. Applying this result to the compact metric space $K=\overline{T(A)}^{\,Y}$ and the sequence $y_k:=Tx_k$ proves that $(Tx_k)$ has a subsequence converging in $Y$.[/guided]

[step:Show the subsequence property forces total boundedness of every image]Assume that every bounded sequence $(x_k)_{k\in\mathbb N}$ in $X$ has the property that $(Tx_k)_{k\in\mathbb N}$ admits a subsequence converging in $Y$. Let $A\subset X$ be bounded, and define \begin{align*} M:=\overline{T(A)}^{\,Y}\subset Y. \end{align*} We prove that $M$ is totally bounded. Fix $\varepsilon>0$. For $y\in Y$ and $r>0$, define the open ball in $Y$ by \begin{align*} B_Y(y,r):=\{z\in Y:\|z-y\|_Y<r\}. \end{align*} Suppose, toward a contradiction, that $M$ cannot be covered by finitely many open balls in $Y$ of radius $\varepsilon$. We construct a sequence $(y_k)_{k\in\mathbb N}$ in $M$ such that \begin{align*} \|y_j-y_k\|_Y\geq \varepsilon \end{align*} whenever $j\neq k$. Choose $y_1\in M$. Having chosen $y_1,\dots,y_k$, the finite union \begin{align*} \bigcup_{i=1}^k B_Y(y_i,\varepsilon) \end{align*} does not cover $M$, so choose \begin{align*} y_{k+1}\in M\setminus \bigcup_{i=1}^k B_Y(y_i,\varepsilon). \end{align*} Then $\|y_{k+1}-y_i\|_Y\geq\varepsilon$ for $1\leq i\leq k$. For each $k\in\mathbb N$, because $y_k\in\overline{T(A)}^{\,Y}$, choose $a_k\in A$ such that \begin{align*} \|Ta_k-y_k\|_Y<\frac{\varepsilon}{4}. \end{align*} The sequence $(a_k)_{k\in\mathbb N}$ is bounded in $X$ because $A$ is bounded. By hypothesis, there is a subsequence $(Ta_{k_j})_{j\in\mathbb N}$ converging in $Y$. Every convergent sequence in a normed space is Cauchy, so there exists $J\in\mathbb N$ such that for all $p,q\geq J$, \begin{align*} \|Ta_{k_p}-Ta_{k_q}\|_Y<\frac{\varepsilon}{2}. \end{align*} For distinct $p,q\geq J$, the triangle inequality gives \begin{align*} \|y_{k_p}-y_{k_q}\|_Y &\leq \|y_{k_p}-Ta_{k_p}\|_Y+\|Ta_{k_p}-Ta_{k_q}\|_Y+\|Ta_{k_q}-y_{k_q}\|_Y \\ &< \frac{\varepsilon}{4}+\frac{\varepsilon}{2}+\frac{\varepsilon}{4} =\varepsilon, \end{align*} contradicting the construction of $(y_k)$. Hence $M$ is totally bounded.[/step]

[guided]Now assume the sequential hypothesis: every bounded sequence in $X$ has an image sequence under $T$ with a convergent subsequence in $Y$. We want to prove compactness of $T$, so take an arbitrary bounded subset $A\subset X$ and set \begin{align*} M:=\overline{T(A)}^{\,Y}\subset Y. \end{align*} The goal is to show that $M$ is compact. The first ingredient is [total boundedness](/page/Total%20Boundedness): for every $\varepsilon>0$, the set $M$ should be coverable by finitely many open $Y$-balls of radius $\varepsilon$. Fix $\varepsilon>0$. For $y\in Y$ and $r>0$, define \begin{align*} B_Y(y,r):=\{z\in Y:\|z-y\|_Y<r\} \end{align*} to be the open ball in $Y$ with centre $y$ and radius $r$. Suppose $M$ is not coverable by finitely many open balls of radius $\varepsilon$. Then we can build points in $M$ that stay $\varepsilon$ apart from all earlier choices. Choose any $y_1\in M$. If $y_1,\dots,y_k$ have been chosen, then \begin{align*} \bigcup_{i=1}^k B_Y(y_i,\varepsilon) \end{align*} is a finite union of open balls of radius $\varepsilon$, and by the failure of finite cover it cannot cover $M$. Hence we can choose \begin{align*} y_{k+1}\in M\setminus \bigcup_{i=1}^k B_Y(y_i,\varepsilon). \end{align*} This guarantees \begin{align*} \|y_{k+1}-y_i\|_Y\geq \varepsilon \end{align*} for each $1\leq i\leq k$. Therefore the whole sequence $(y_k)$ is pairwise $\varepsilon$-separated. The points $y_k$ may lie in the closure of $T(A)$ rather than in $T(A)$ itself, so we approximate them by actual images. Since $y_k\in\overline{T(A)}^{\,Y}$, the open ball $B_Y(y_k,\varepsilon/4)$ intersects $T(A)$. Thus for each $k\in\mathbb N$ there is $a_k\in A$ such that \begin{align*} \|Ta_k-y_k\|_Y<\frac{\varepsilon}{4}. \end{align*} Because $A$ is bounded, the sequence $(a_k)_{k\in\mathbb N}$ is bounded in $X$. The sequential hypothesis applies and gives a subsequence $(Ta_{k_j})_{j\in\mathbb N}$ converging in $Y$. A convergent sequence in a normed space is Cauchy. Hence there exists $J\in\mathbb N$ such that whenever $p,q\geq J$, \begin{align*} \|Ta_{k_p}-Ta_{k_q}\|_Y<\frac{\varepsilon}{2}. \end{align*} For distinct $p,q\geq J$, the triangle inequality gives \begin{align*} \|y_{k_p}-y_{k_q}\|_Y &\leq \|y_{k_p}-Ta_{k_p}\|_Y+\|Ta_{k_p}-Ta_{k_q}\|_Y+\|Ta_{k_q}-y_{k_q}\|_Y \\ &< \frac{\varepsilon}{4}+\frac{\varepsilon}{2}+\frac{\varepsilon}{4} =\varepsilon. \end{align*} This contradicts the pairwise separation $\|y_{k_p}-y_{k_q}\|_Y\geq\varepsilon$. Therefore our supposition was false, and $M=\overline{T(A)}^{\,Y}$ is totally bounded.[/guided]

[step:Use completeness to turn total boundedness into compactness]Let $A\subset X$ be bounded and let \begin{align*} M:=\overline{T(A)}^{\,Y}. \end{align*} The previous step shows that $M$ is totally bounded. Since $Y$ is complete and $M$ is closed in $Y$, the metric space $M$ is complete. Define the inherited metric \begin{align*} d_M:M\times M&\to [0,\infty) \\ (z,w)&\mapsto \|z-w\|_Y, \end{align*} and for $z\in M$ and $r>0$ define the open ball in $M$ by \begin{align*} B_M(z,r):=\{w\in M:d_M(w,z)<r\}. \end{align*} We prove that a complete totally bounded metric space is compact by first proving sequential compactness and then proving that sequential compactness implies open-cover compactness in this metric setting. Let $(z_k)_{k\in\mathbb N}$ be a sequence in $M$. For each $m\in\mathbb N$, total boundedness gives a finite cover of $M$ by open balls of radius \begin{align*} \frac{1}{m}. \end{align*} Choose inductively nested infinite index sets \begin{align*} N_1\supset N_2\supset \cdots \subset \mathbb N \end{align*} such that $\{z_k:k\in N_m\}$ lies in one open ball of radius $1/m$. Then choose indices $k_m\in N_m$ inductively with \begin{align*} k_1<k_2<\cdots. \end{align*} This is possible because each $N_m$ is infinite, so after $k_{m-1}$ has been chosen there exists an element of $N_m$ larger than $k_{m-1}$. For $p,q\geq m$, both $k_p$ and $k_q$ belong to $N_m$, so $z_{k_p}$ and $z_{k_q}$ lie in the same ball of radius $1/m$. Therefore \begin{align*} d_M(z_{k_p},z_{k_q})<\frac{2}{m}, \end{align*} Thus $(z_{k_m})_{m\in\mathbb N}$ is Cauchy in $M$, and completeness of $M$ gives convergence in $M$. Hence every sequence in $M$ has a convergent subsequence. To prove compactness, suppose an open cover $\mathcal U$ of $M$ has no finite subcover. Since $M$ is totally bounded, choose a finite cover of $M$ by open balls of radius $1$. At least one ball has closure in $M$ that is not finitely covered by $\mathcal U$; otherwise finitely many finite subcovers would cover $M$. Call this closed subset $F_1\subset M$. Having chosen a closed subset $F_m\subset M$ that is not finitely covered by $\mathcal U$, cover $F_m$ by finitely many open balls in $M$ of radius \begin{align*} \frac{1}{m+1}. \end{align*} At least one of the relative closures of the intersections of these balls with $F_m$ is not finitely covered by $\mathcal U$; choose it and call it $F_{m+1}$. This gives nested nonempty closed subsets \begin{align*} F_1\supset F_2\supset \cdots \end{align*} not finitely covered by $\mathcal U$, with $F_m$ contained in an open ball of radius \begin{align*} \frac{1}{m}. \end{align*} Choose $w_m\in F_m$ for each $m\in\mathbb N$. By the sequential compactness just proved, a subsequence $(w_{m_j})_{j\in\mathbb N}$ converges to some point $z\in M$. Since $\mathcal U$ covers $M$, choose $U\in\mathcal U$ with $z\in U$. The openness of $U$ gives $\rho>0$ such that $B_M(z,\rho)\subset U$. Choose $j\in\mathbb N$ so large that \begin{align*} \frac{2}{m_j}<\frac{\rho}{3} \end{align*} and $d_M(w_{m_j},z)<\rho/3$. Since $F_{m_j}$ is contained in an open ball of radius $1/m_j$ and contains $w_{m_j}$, every $w\in F_{m_j}$ satisfies \begin{align*} d_M(w,z) &\leq d_M(w,w_{m_j})+d_M(w_{m_j},z) \\ &< \frac{2}{m_j}+\frac{\rho}{3} <\rho. \end{align*} Thus $F_{m_j}\subset U$, contradicting that $F_{m_j}$ is not finitely covered by $\mathcal U$. Thus every open cover of $M$ has a finite subcover, so $M$ is compact. Therefore $\overline{T(A)}^{\,Y}$ is compact for every bounded subset $A\subset X$. This is exactly the compactness of $T$.[/step]

[guided]We now finish the reverse implication. Let $A\subset X$ be bounded and define \begin{align*} M:=\overline{T(A)}^{\,Y}. \end{align*} From the previous step, $M$ is totally bounded. Since $Y$ is complete and $M$ is closed in $Y$, every [Cauchy sequence](/page/Cauchy%20Sequence) in $M$ converges in $Y$ to a point of $M$; hence $M$ is complete. Define the inherited metric \begin{align*} d_M:M\times M&\to [0,\infty) \\ (z,w)&\mapsto \|z-w\|_Y, \end{align*} and for $z\in M$ and $r>0$ define \begin{align*} B_M(z,r):=\{w\in M:d_M(w,z)<r\} \end{align*} to be the open ball in $M$ with centre $z$ and radius $r$. Why does completeness matter? Total boundedness alone gives finite approximations at every scale, but it does not guarantee that limiting points remain in the space. Completeness supplies those limits. We first show that every sequence in $M$ has a convergent subsequence. Let $(z_k)_{k\in\mathbb N}$ be a sequence in $M$. For $m=1$, total boundedness gives finitely many balls of radius $1$ covering $M$, so one of them contains infinitely many terms of the sequence. Let $N_1\subset\mathbb N$ be an infinite set of indices for those terms. Repeating the same argument inside the infinite set $N_1$, total boundedness gives a ball of radius $1/2$ containing infinitely many of the terms indexed by $N_1$; call that infinite index set $N_2$. Inductively, we obtain nested infinite sets \begin{align*} N_1\supset N_2\supset \cdots \end{align*} such that $\{z_k:k\in N_m\}$ lies in one open ball of radius $1/m$. Choose indices $k_m\in N_m$ inductively with \begin{align*} k_1<k_2<\cdots. \end{align*} This choice is possible because each $N_m$ is infinite, so after $k_{m-1}$ has been chosen there is an element of $N_m$ larger than $k_{m-1}$. If $p,q\geq m$, then $k_p,k_q\in N_m$, so $z_{k_p}$ and $z_{k_q}$ lie in the same ball of radius $1/m$. Hence \begin{align*} d_M(z_{k_p},z_{k_q})<\frac{2}{m}, \end{align*} This proves that $(z_{k_m})$ is Cauchy. Since $M$ is complete, this subsequence converges to a point of $M$. It remains to prove, in this metric setting, that the sequential compactness just obtained implies open-cover compactness. Suppose, for contradiction, that some open cover $\mathcal U$ of $M$ has no finite subcover. Since $M$ is totally bounded, finitely many open balls in $M$ of radius $1$ cover $M$. At least one of their relative closures in $M$ is not finitely covered by $\mathcal U$; otherwise finitely many finite subcovers would cover $M$. Call such a nonempty closed subset $F_1\subset M$. Assume that a nonempty closed subset $F_m\subset M$ has been chosen and is not finitely covered by $\mathcal U$. Cover $F_m$ by finitely many open balls in $M$ of radius \begin{align*} \frac{1}{m+1}. \end{align*} Among the relative closures in $M$ of the intersections of these balls with $F_m$, at least one is not finitely covered by $\mathcal U$; choose it and call it $F_{m+1}$. This produces nested nonempty closed subsets \begin{align*} F_1\supset F_2\supset\cdots \end{align*} such that each $F_m$ is not finitely covered by $\mathcal U$ and $F_m$ is contained in an open ball in $M$ of radius $1/m$. Choose $w_m\in F_m$ for each $m\in\mathbb N$. By sequential compactness, a subsequence $(w_{m_j})_{j\in\mathbb N}$ converges to some $z\in M$. Since $\mathcal U$ covers $M$, choose $U\in\mathcal U$ with $z\in U$. Because $U$ is open in $M$, there exists $\rho>0$ such that \begin{align*} B_M(z,\rho)\subset U. \end{align*} Choose $j\in\mathbb N$ so large that \begin{align*} \frac{2}{m_j}<\frac{\rho}{3} \end{align*} and $d_M(w_{m_j},z)<\rho/3$. Since $F_{m_j}$ is contained in an open ball of radius $1/m_j$ and contains $w_{m_j}$, every $w\in F_{m_j}$ satisfies \begin{align*} d_M(w,z) &\leq d_M(w,w_{m_j})+d_M(w_{m_j},z) \\ &< \frac{2}{m_j}+\frac{\rho}{3} <\rho. \end{align*} Thus $F_{m_j}\subset U$, contradicting that $F_{m_j}$ is not finitely covered by $\mathcal U$. Hence every open cover of $M$ has a finite subcover, and $M$ is compact. Since $A\subset X$ was an arbitrary bounded subset, $\overline{T(A)}^{\,Y}$ is compact for every bounded $A$. This is precisely the definition that $T$ is compact.[/guided]