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A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $[0,1]$, and letThe set is nonempty. If $s=\sup A<1$, choose $U\in \mathcal{U}$ containing $s$. An interval about $s$ lies in $U$; a point of $A$ just to the left of $s$ then extends the finite cover beyond $s$, contradicting the definition of the supremum. Hence $s=1$, and the same neighbourhood argument covers $1$, proving $[0,1]$ compact.

The continuous map $t\mapsto (\cos 2\pi t,\sin 2\pi t)$ maps $[0,1]$ onto $S^1$. A continuous image of a compact space is compact: pull an open cover back, choose a finite subcover, and push the corresponding sets forward. Thus $S^1$ is compact.

In the unusual topology on $X={\mathbb{R}}^2\setminus \{0\}$, a proper closed set is a finite union of punctured lines through the origin. Take any nonempty member $U$ of an open cover. Its complement is such a finite union. Each punctured line has the indiscrete subspace topology, because every other vector subspace meets it only at the removed origin, and is therefore compact. Finitely many finite subcovers, together with $U$, cover $X$. Hence $X$ is compact.

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.

No. Give an infinite set $X$ the discrete metricThis space is bounded. Every Cauchy sequence is eventually constant, so it is complete. However, the open cover by the singleton balls $B_{1/2}(x)=\{x\}$ has no finite subcover. Hence $X$ is not compact.

A filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes the empty set, is upward closed, and is closed under finite intersections. An ultrafilter is a proper filter that contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To prove the ultrafilter lemma, order the proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is again a proper filter, so Zorn lemma gives a maximal extension $\mathcal{U}$. If neither $A$ nor $X\setminus A$ belonged to $\mathcal{U}$, adjoining either one would generate an improper filter. There would then be $U,V\in \mathcal{U}$ with $U\cap A=\varnothing$ and $V\cap (X\setminus A)=\varnothing$. But $U\cap V=\varnothing$, contradicting propriety. Hence $\mathcal{U}$ is an ultrafilter.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters on $\mathbb{N}$, with basic setsThe identitiesshow that these sets form a basis of clopen sets. Distinct ultrafilters disagree on some $A$; one lies in $\overline{A}$ and the other in the disjoint set $\overline{\mathbb{N}\setminus A}$. Thus $\beta \mathbb{N}$ is a Hausdorff space.

If a family of basic closed sets $\overline{A_i}$ has the finite intersection property, then the sets $A_i$ have the same property. They generate a proper filter, which extends to an ultrafilter lying in every $\overline{A_i}$. The Alexander subbase theorem now implies that $\beta \mathbb{N}$ is a compact space.

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzela-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.

Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the open set $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a compact space is compact.

Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint open sets $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. PutThen $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Thereforeare disjoint open neighbourhoods of $A$ and $B$. This proves the normality of a compact Hausdorff space.

Let ${\rho }_{n,j}^{\prime }$ be the $j$th one-system marginal of ${\rho }_n^{\prime }$, and put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1$. Contractivity of trace distance under partial trace givesuniformly in $j$. Because the one-system state space is a compact space, continuity of $f(\cdot \Vert \sigma )$ implies uniform continuity. Hence there is a function $\delta (\epsilon )\to 0$ such thatfor every $j$.

Multipartite superadditivity and additivity now implyThereforewhich proves the required lower asymptotic semicontinuity of quantum relative entropy argument.

Choosewhere $r_p$ is a normal radius at $p$. Take piecewise smooth curves $c_j$ from $p$ to $q$ such that $L_g(c_j)\to d_g(p,q)$. Each $c_j$ first leaves the normal ball $B_g(p,\delta )$ at a point $p_j$. The Gauss lemma gives $d_g(p,p_j)=\delta$. The geodesic sphereis compact, because the tangent-space sphere is compact and ${\exp }_p$ is defined on it. After taking a convergent subsequence, let $p_j\to p_0$.

The part of $c_j$ after $p_j$ has length at least $d_g(p_j,q)$, so continuity of the Riemannian distance givesThe triangle inequality gives the reverse inequality. Therefore

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.

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.