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Every continuous map $T:X\to X$ on a sequentially compact space with a metric satisfying $d(Tx,Ty)\ge d(x,y)$ is a surjective isometry.

A convergent subsequence of the paired orbit $(T^{k}x,T^{k}y)$ has two late terms arbitrarily close. Removing their common number of iterates by the distance inequality shows that $(x,y)$ returns arbitrarily close to itself after a common positive time $n$. Thus $d(Tx,Ty)\le d(T^{n}x,T^{n}y)\le d(x,y)+2ϵ$, and letting $ϵ\downarrow 0$ proves the isometry claim. For each $x$, choose return times $n_k\ge 1$ with $T^{n_k}x\to x$. A convergent subsequence of $T^{n_k-1}x$ has limit $z$, and continuity gives $Tz=x$. This proves surjectivity.

Suppose the continuous $f$ were not uniformly continuous. There would be an ${\epsilon }_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. Sequential compactness gives a subsequence $x_{n_j}\to x\in X$. The triangle inequality gives $y_{n_j}\to x$ as well. Continuity at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to 0$, a contradiction. Thus $f$ is uniformly continuous.

It need not be Lipschitz continuous. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt{x}$ hasNo finite Lipschitz constant exists.

A compact space is one for which every open cover has a finite subcover. A metric space is sequentially compact when every sequence in it has a convergent subsequence whose limit belongs to the space.

Let $(x_n)$ be any sequence in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The diagonal argument produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hencewhenever $k,\ell \ge j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.

Every sequence in $X$ therefore has a convergent subsequence, so $X$ is sequentially compact. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a complete totally bounded metric space is compact.

The assumptions prove convergence to the set of J-minimizing solutions, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.

Let $u^{†}$ be any J-minimizing solution, whose existence is granted. It is feasible for every noisy-data constraint because $A{u}^{†}=f$ and $\Vert f-f_{\delta }\Vert \le \delta$. Minimality therefore gives $J(u_{\delta })\le J(u^{†})$. All reconstructions lie in this strongly sequentially compact space sublevel set. For any sequence ${\delta }_k\downarrow 0$, pass to a subsequence with $u_{{\delta }_k}\to u_{\ast }$. Boundedness of $A$ and the triangle inequality implyso $A{u}_{\ast }=f$. By sequential lower semicontinuity,Thus $u_{\ast }$ is a J-minimizing solution. The same compactness argument shows the distance to a set of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full strong convergence.

For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $A=0$, $f=f_{\delta }=0$, and $J(u)=\max \{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\ge 0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1{)}^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is compact-sublevel convergence of the residual method.

Assume for contradiction that $X$ is infinite-dimensional. Starting with $M_0=\{0\}$, apply Riesz lemma recursively to the proper closed finite-dimensional subspacesto choose unit vectors $x_n$ satisfyingFor $m<n$, the vector $x_m$ belongs to $M_{n-1}$, and thereforeThus $(x_n)$ lies in the closed unit ball but has no Cauchy subsequence, hence no convergent subsequence. A compact metric space is sequentially compact, contradicting compactness of the unit ball. Therefore $X$ is finite-dimensional, proving that a compact unit ball characterizes finite-dimensional normed spaces.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.

For metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, the sum metric on $X_1\times X_2$ is the displayed formula. Nonnegativity, symmetry and the triangle inequality follow by adding the respective inequalities in the two factors, and zero distance forces equality in both coordinates. It induces the finite product topology, as does the maximum product metric, since their distances satisfy $d_{\max }\le d_{\mathrm{sum}}\le 2d_{\max }$. A sequence converges for this metric exactly when both coordinate sequences converge. Two successive extractions of a convergent subsequence therefore show that the product of two sequentially compact spaces is sequentially compact.