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An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

For fixed $x\in X$, the function $a\mapsto d(x,a)$ on $A$ is continuous, indeed $1$-Lipschitz continuous by the reverse triangle inequality. A continuous real-valued function on a nonempty compact space attains its minimum by the extreme value theorem. Hence some $a_0\in A$ satisfiesThe metric is symmetric, so this is also $d(a_0,x)$. Compactness supplies attainment of the infimum.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

A subset $K$ of a topological space is compact if every cover of $K$ by open subsets of the ambient space has a finite subcover. This is equivalent to compactness of $K$ with its subspace topology.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

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.

Let $(V_i)$ be an open cover of $Y$. By continuity, the inverse images $f^{-1}(V_i)$ are open in $X$; by surjectivity, they cover $X$. Compactness supplies finitely many covering $X$. Applying $f$ and using $f(X)=Y$ shows that the corresponding finitely many $V_i$ cover $Y$.

Thus a continuous surjective function from a compact space has compact target. This is the continuous image of a compact space theorem.

Let $F$ be a closed subset of the compact space $X$. Express an arbitrary open cover of $F$ in its subspace topology as $\{F\cap U_i:i\in I\}$, where the $U_i$ are open sets of $X$. Since $X\setminus F$ is open, the familyis an open cover of $X$. Take a finite subcover and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded set contains no point of $F$. Their traces on $F$ give the required finite subcover. Hence every closed subset of a compact space is compact, with no Hausdorff assumption.

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

Write $K={\bigcap }_{i\in I}{K}_i$. Each $K_i$ is closed in $X$, by the compact subset of a Hausdorff space theorem. Therefore $K$ is closed, and is compact because $X$ is compact. The nonempty finite intersections give the finite intersection property, so part (a)(ii) also yields $K\ne \varnothing$.

Suppose $K$ were not connected. Part (c) would give disjoint open sets $U,V$ containing $K$ in their union and each meeting $K$. LetThis is a closed subset of the compact space $X$, and $C\cap {\bigcap }_{i\in I}{K}_i=\varnothing$. Apply the contrapositive of the finite intersection characterization of compactness to the family of closed sets $\{C\}\cup \{K_i:i\in I\}$. Some finite subfamily has empty intersection. Since every finite intersection of just the $K_i$ is nonempty, that subfamily must include $C$. Thus for some finite $J\subseteq I$,But $K\subseteq K_J$ and $K$ meets both $U$ and $V$. Therefore $K_J\cap U$ and $K_J\cap V$ are nonempty disjoint relatively open sets covering $K_J$, contradicting its assumed connectedness. This provesThis is connectedness of an intersection from connected finite intersections. If $I=\varnothing$, its intersection is $X$; the stated hypothesis for the finite set $J=\varnothing$ already makes $X$ nonempty and connected, so that case is included.

The product topology of two compact spaces is compact, without a Hausdorff condition. Refine an open cover locally by basis rectangles. For fixed $a\in A$, compactness of $B$ selects finitely many rectangles whose second factors cover $B$. Their first factors have an open finite intersection $U_a$ containing $a$, and the finitely many original covering members cover $U_a\times B$. Use compactness of $A$ to choose finitely many $U_a$ covering $A$. The associated finitely many original members form a finite subcover of $A\times B$. Repeating the argument proves the finite-factor case of Tychonoff's theorem.

For locally compact noncompact Hausdorff spaces, a continuous proper map $f:X\to Y$ extends continuously to their one-point compactifications by the displayed rule. The inverse image of $\alpha Y\setminus K$ is $\alpha X\setminus f^{-1}(K)$, a neighborhood of infinity because the inverse image of a compact set is compact. The extension itself is proper, since it is continuous from a compact space to a Hausdorff space. A nonproper map can fail to admit any continuous extension: projection ${\mathbb{R}}^2\to \mathbb{R}$ sends two vertical sequences escaping to the same point at infinity to two distinct constant sequences, incompatible with continuity at that point.

The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

An arbitrary product of compact spaces is compact in the product topology. The theorem is equivalent to the axiom of choice.

A topological space is both a compact space and a Hausdorff space exactly when every ultrafilter on its underlying set converges to exactly one point. Existence of all ultrafilter limits characterizes compactness. If a space is not Hausdorff, two distinct points have neighbourhood systems whose combined finite intersections are nonempty; the ultrafilter lemma extends these to an ultrafilter converging to both points. Conversely, disjoint neighbourhoods prevent such a common limit.

A point $x$ is the limit of a map $f:I\to X$ along an ultrafilter $\mathcal{U}$ on $I$ when $f^{-1}(V)\in \mathcal{U}$ for every neighbourhood $V$ of $x$. Every ultrafilter on a compact space has a limit, and that limit is unique in a Hausdorff space.

The same Banach space can acquire incomparable weak-star topologies from different isometric preduals. Pair ${\ell }^1$ with $c_0$ by $\sum {\alpha }_n{\beta }_n$ and with the space $c$ of convergent sequences by ${\alpha }_1{\gamma }_{\infty }+{\sum }_{n\ge 2}{\alpha }_n{\gamma }_{n-1}$. The functional $\alpha \mapsto {\alpha }_1$ is continuous only for the first topology: $e_1-e_n\to 0$ for the second but its values remain one. Conversely $\alpha \mapsto \sum {\alpha }_n$ is continuous only for the second: $e_n\to 0$ for the first but its values remain one. More generally, comparable weak-star topologies arising from isometric preduals agree on the common closed unit ball, since its identity map is a continuous bijection from a compact space to a Hausdorff space by the Banach-Alaoglu theorem.