Let $V$ be a neighbourhood of $A$ and suppose it possible that every set of $\mathfrak{F}$ meets $\complement V$. The intersections of the sets of $\mathfrak{F}$ with $\complement V$ then form a base of a filter $\mathfrak{G}$ on $X$; $X$ is quasi-compact, so that $\mathfrak{G}$ has at least one cluster point $y$, which does not belong to $A$, because the neighbourhood $V$ of $A$ does not meet some of the sets of $\mathfrak{G}$. But since $\mathfrak{G}$ is finer than $\mathfrak{F}$, $y$ is also a cluster point of $\mathfrak{F}$, which is contrary to hypothesis.

Necessity by § 8, no. 1, Proposition 1; sufficiency by Theorem 1 above.

The condition is necessary since X is Hausdorff (§ 8, no. 1, Proposition 1). It is sufficient, for it signifies that x is the only cluster point of B; hence B converges to x by the Corollary to Theorem 1 of no. 1.

For it follows from axiom (Hi) (§ 8, no. 1, Proposition 1) that the filter base formed by all the closed neighbourhoods of an arbitrary point of the space satisfies the condition of Proposition 1.

The following proposition amplifies the Corollary to Proposition 1:

Suppose the conclusion is false. If every neighbourhood U of A meets every neighbourhood V of B, then the sets U ∩ V form a filter base B on X, which therefore has a cluster point x ∈ X. Now x must lie in A, since if y is any point of X not in A there is a neighbourhood of y and a neighbourhood of A which do not intersect, since X is regular, and thus y cannot be a cluster point of B. Similarly x must lie in B and we have a contradiction.

This proposition has important consequences which will be examined in Chapter IX, § 4.

The non-Hausdorff quasi-compact space X of Example 2 of no. 1 does not satisfy axiom (O_{III}), nor a fortiori the property stated in Proposition 2, since any two non-empty open sets of this space always intersect.

This is an immediate consequence of axiom $(C'')$ if we remark that if $A$ is closed in $X$ then every set which is closed in $A$ is closed in $X$.

Let $A$ be a compact subset of a Hausdorff space $X$, and let $x$ be any point of $\overline{A}$; we have to show that $x \in A$. By hypothesis, every neighbourhood of $x$ meets $A$, and therefore the neighbourhood filter $\mathfrak{B}$ of $x$ in $X$ induces a filter $\mathfrak{B}_A$ on $A$; $A$ is compact whence $\mathfrak{B}_A$ has a cluster point $y \in A$. Since the filter $\mathfrak{B}$ is coarser than the filter on $X$ generated by $\mathfrak{B}_A$ (considered as a filter base on $X$), $y$ is also a cluster point of $\mathfrak{B}$; hence $y = x$, because $\mathfrak{B}$ converges to $x$ in $X$ and $X$ is Hausdorff (§ 8, no. 1, Proposition 1).

It is sufficient to show that if $A$ and $B$ are two quasi-compact subsets of a topological space $X$, then $A \cup B$ is quasi-compact. Let $\mathscr{R}$ be covering of $A \cup B$; then $\mathscr{R}$ is a covering of $A$ and a covering of $B$; hence $\mathscr{R}$ contains a finite covering $\mathscr{R}_1$ of $A$ and a finite covering $\mathscr{R}_2$ of $B$; $\mathscr{R}_1 \cup \mathscr{R}_2$ is thus a finite covering of $A \cup B$ contained in $\mathscr{R}$.

If $A$ is relatively compact, then $\overline{A}$ is compact by Proposition 4 and its corollary; the reverse implication is self-evident.

For if $A \subset K$, where $K$ is a quasi-compact subset of $X$, then every filter base on $A$ has a cluster point in $K$.

The converse of this proposition is not valid without restriction on $X$ (Exercise 22).

Let $\mathscr{R}$ be a covering of $f(X)$ by open sets in $X'$; then $\overline{f}^{-1}(\mathscr{R})$ is an open covering of $X$ (§ 2, no. 1, Theorem 1); hence there is a finite subset $\mathscr{S}$ of $\mathscr{R}$ such that $(\mathscr{S} \overline{f})$ is a covering of $X$; but then $\mathscr{S}$ is a covering of $f(X)$ and the theorem is proved.

This follows immediately from Corollary 1 and Proposition 4 of no. 3.

In particular:

a) If there is a quasi-compact set $K$ in $X$ which meets every equivalence class mod $R$, then $X/R$ is compact and the canonical mapping of $K/R_K$ onto $X/R$ is a homeomorphism.

b) If $K$ also meets each equivalence class in only one point, then $K$ is a continuous section of $X$ with respect to the relation $R$ (§ 3, no. 5).

Let $f$ be the restriction to $K$ of the canonical mapping $X \to X/R$. Since $X/R$ is Hausdorff it follows from Corollary 1 that $X/R$ is compact and from Corollary 2 that $f$ is closed; hence the bijection $K/R_K \to X/R$ associated with $f$ is a homeomorphism (§ 5, no. 2, Proposition 3). This deals with a); b) follows immediately, since we now have $K/R_K = K$.

In view of the characterization of Hausdorff product spaces given in § 8, no. 2, Proposition 7 it is enough to prove the assertions for quasi-compact spaces. If $X = \prod_{i \in I} X_i$ is quasi-compact and non-empty, then $X_i = \operatorname{pr}_i(X)$ is quasi-compact by reason of Theorem 2 of no. 4. Conversely, suppose the $X_i$ are quasi-compact and let $\mathfrak{U}$ be an ultrafilter on $X$; then for each $i \in I$, $\operatorname{pr}_i(\mathfrak{U})$ is an ultrafilter base on $X_i$ (§ 6, no. 6, Proposition 10) which therefore converges by reason of axiom (C'); hence $\mathfrak{U}$ is convergent (§ 7, no. 6, Corollary 1 of Proposition 10) and therefore $X$ is quasi-compact.

Necessity follows from Theorem 2 of § 4. To prove sufficiency, let $A$ be a subset of $\prod_i X_i$ such that, for each index $i$, $\operatorname{pr}_i(A)$ is contained in a quasi-compact subset $K_i$ of $X_i$; then $A$ is contained in the quasi-compact subset $\prod_i K_i$ of $\prod_i X_i$.

$X$ is a closed subspace of $\prod_\alpha X_\alpha$ (§ 8, no. 2, Proposition 7, Corollary 2) which is compact by Theorem 3 of no. 5 and Proposition 3 of no. 3. The other assertions are consequences of Set Theory, chap. III, § 7, no. 4, th. 1. We apply this theorem by taking $\mathscr{S}_\alpha$ to be the set of closed subsets of $X_\alpha$. The conditions (i) and (ii) are just axioms $(O'_1)$ and $(C'')$ respectively; condition (iii) is satisfied since $\{x_\alpha\}$ is closed and $f_{\alpha\beta}$ continuous (§ 2, no. 1, Theorem 1), and lastly condition (iv) is satisfied by reason of Corollary 2 of Theorem 2 of no. 4.

For each $x_\alpha \in \bigcap_{\beta \geq \alpha} f_{\alpha\beta}(X_\beta)$ and each $\beta \geq \alpha$, let $L_\beta$ denote $\overline{f}_{\alpha\beta}^{-1}(x_\alpha)$. If $\alpha \leq \beta \leq \gamma$, then we have $f_{\beta\gamma}(L_\gamma) \subset L_\beta$ and the set of all indices $\beta \geq \alpha$ is cofinal in the index set. It follows immediately that the $L_\beta$ ($\beta \geq \alpha$) form an inverse system of topological spaces (with the restrictions of the $f_{\beta\gamma}$ as mappings), whose inverse limit $L$ is homeomorphic to $\overline{f}_{\alpha}^{-1}(x_\alpha)$. Since by hypothesis the $L_\beta$ are compact and not empty, the corollary follows from Proposition 8.

Let $L_\alpha$ denote $\overline{u}_\alpha^{-1}(x'_\alpha)$; then clearly the $L_\alpha$ form an inverse system of topological spaces (with the restrictions of the $f_{\alpha\beta}$ as maps) and $\overline{u}^{-1}(x' = L)$ is the inverse limit of the $L_\alpha$; assertion a) therefore follows from Proposition 8. Assertion b) is an immediate consequence, in view of Proposition 3 of no. 3.

Let $X$ be a locally compact space, then every point $x \in X$ has a compact neighbourhood $V$; since $X$ is Hausdorff, $V$ is closed (no. 3, Proposition 4). On the other hand, $V$ is a regular subspace of $X$ (no. 2, Corollary to Proposition 1) and therefore $X$ is regular (§ 8, no. 4, Proposition 13).

For the intersection of a closed neighbourhood of $x$ and a compact neighbourhood of $x$ is a compact neighbourhood of $x$ (no. 3, Proposition 3).

There exist non-Hausdorff topological spaces in which every point has a fundamental system of compact neighbourhoods (Exercise 5).

The Corollary to Proposition 9 may be generalized as follows:

Let $U$ be any neighbourhood of $K$. For each $x \in K$ there is a compact neighbourhood $W(x)$ of $x$ contained in $U$. The interiors of the sets

W(x) form an open covering of K as x runs through K; hence there exist a finite number of points $x_i \in K$ ($1 \leq i \leq n$) such that the interiors of the W(x_i) cover K. The union V of the W(x_i) is therefore a compact neighbourhood of K contained in U (no. 3, Proposition 5).

In view of Proposition 4 of no. 3, this follows from Proposition 3 a) of § 3, no. 1.

By hypothesis, for every $x \in A$ there is a neighbourhood V of x in X such that $V \cap A$ is compact and therefore closed in V (no. 3, Proposition 4).

Let A be a locally closed subspace of X; then for each $x \in A$ there is a neighbourhood U of x in X such that $U \cap A$ is closed in U. Let $V \subset U$ be a compact neighbourhood of x in X; $V \cap A = V \cap (U \cap A)$ is closed in V and therefore compact (no. 3, Proposition 3). Since $V \cap A$ is a neighbourhood of x in A, the result is proved (it is clear that A is Hausdorff).

Theorem 1 (no. 1) and Corollary 2 of Theorem 2 (no. 4) do not extend to locally compact spaces which are not compact.

For example, in an infinite discrete space X, the filter consisting of those sets which contain a given point $x \in X$ and have a finite complement has x as its only cluster point but does not converge to x. Since any mapping f of X into a Hausdorff space X' is continuous, the image under f of an arbitrary subset of X (which is closed in X, since X is discrete) will not in general be a closed subset of X'.

The proposition corresponding to Theorem 3 of no. 5 is the following:

b) Conversely, if the product of a family $(X_i)_{i \in I}$ of non-empty topological spaces is locally compact, then the factors $X_i$ are compact for all but a finite number of indices, and the factors which are not compact are locally compact.

a) Let $x = (x_i)$ be a point of $X$. For each index $i \in I$ such that $X_i$ is locally compact but not compact, let $V_i$ be a compact neighbourhood of $x_i$ in $X_i$, and for all other indices $i$ put $V_i = X_i$. Then $\prod_{i \in I} V_i$ is a compact neighbourhood of $x$ in $X$ (no. 5, Theorem 3). Also $X$ is Hausdorff by , no. 2, Proposition 7, and is therefore locally compact

b) If $X = \prod_{i \in I} X_i$ is locally compact and the $X_i$ non-empty, then each of the $X_i$ is homeomorphic to a closed subspace of $X$ ($§ 4$, no. 2, Proposition 4 and $§ 4$, no. 3, Corollary to Proposition 7), hence locally compact by Proposition 13. Let $a = (a_i)$ be a point of $X$ and let $V$ be a compact neighbourhood of $a$; since we have $\operatorname{pr}_i V = X_i$ for all but a finite number of indices ($§ 4$, no. 1), it follows from no. 4, Corollary 1 to Theorem 2, that the $X_i$ are compact except for a finite number of indices.

Let us begin by proving the second assertion of the theorem. Let

If the homeomorphism $g$ exists it must be unique, for by definition we have $g(x') = f_1(\overline{f}(x'))$ if $x' \neq \omega$ and therefore $g(\omega) = \omega_1$. It remains to show that the bijection $g : X' \to X'_1$ thus defined is bicontinuous; since $X'$ and $X'_1$ are interchangeable we need only show that the image under $g$ of a neighbourhood of a point $x' \in X'$ is a neighbourhood of $g(x')$ in $X'_1$. This is obvious from the definition of $g$ if $x' \neq \omega$. If $x' = \omega$, let $V'$ be an open neighbourhood of $\omega$ in $X'$; then $X' - V' = K$ is closed in $X'$ and therefore compact (no. 3, Proposition 3) and is contained in $f(X)$; hence $g(K) = f_1(\overline{f}(K))$ is compact (no. 4, Theorem 2, Corollary 1). It follows that $g(V') = X'_1 - g(K)$ is an open neighbourhood of $\omega_1$ (no. 3, Proposition 4). Hence $g$ is a homeomorphism.

To prove the first part of the theorem, let $X'$ be a set which is the sum of $X$ and a set consisting of a single point $\omega$. We define a topology on $X'$ by taking the set $\mathscr{O}$ of open subsets of $X'$ to consist of all open subsets of $X$ and all subsets of the form $(X - K) \cup \{ \omega \}$, where $K$ is a compact subset of $X$. Since any intersection of compact subsets of $X$ is compact (no. 3, Propositions 3 and 4) and since any closed subset of a compact set is compact (no. 3, Proposition 3), it follows that $\mathfrak{D}$ satisfies axiom $(O_I)$; and since any finite union of compact subsets of $X$ is compact (no. 3, Proposition 5), $\mathfrak{D}$ also satisfies $(O_{II})$. Every compact subset of $X$ is closed in $X$ (no. 3, Proposition 4) and therefore the topology induced on $X$ by that of $X'$ is the original topology on $X$. Thus it remains to show that $X'$ is compact. In the first place, $X'$ is Hausdorff. For if $x, y$ are any two distinct points of $X$, they have disjoint open neighbourhoods $V, W$ respectively in $X$, and $V, W$ are open in $X'$; on the other hand each $x \in X$ has a compact neighbourhood $K$ in $X$, which is also a neighbourhood of $x$ in $X'$, while

is a neighbourhood of $\omega$ in $X'$ and manifestly does not meet $K$. Finally, $X'$ is quasi-compact. Let $(U_\lambda)_{\lambda \in L}$ be an open covering of $X'$; then for at least one index $\mu \in L$ we have $U_\mu = (X - K_\mu) \cup \{ \omega \}$ where $K_\mu$ is a compact subset of $X$. Hence there is a finite subset $H$ of $L$ such that the sets $U_\lambda$ (for $\lambda \in H$) cover $K_\mu$; if $J = H \cup \{ \mu \}$, then $(U_\lambda)_{\lambda \in J}$ is a finite open covering of $X'$, and the proof is complete.

Notice that if $X$ is already compact, then $\omega$ is an isolated point of the compact space $X'$; hence $X'$ is the sum ($\S$ 2, no. 4, Example 3) of the space $X$ and the space $\{ \omega \}$.

When a compact space $X'$ has been constructed as above from a locally compact space $X$ by adjoining an element $\omega$, it is often said that $\omega$ is the "point at infinity" of $X'$, and that $X'$ is obtained from $X$ by adjoining a point at infinity. $X'$ is also called the Alexandroff compactification or the one-point compactification of the locally compact space $X$.

\* Example. If we apply Alexandroff's theorem to the real plane $\mathbf{R}^2$, we get a compact space homeomorphic to the sphere $S_2$ whose equation is $x_1^2 + x_2^2 + x_3^2 = 1$ in $\mathbf{R}^3$. A homeomorphism of these two spaces may be described as follows: the point $\omega$ (the point at infinity) adjoined to $\mathbf{R}^2$ is mapped to $(0, 0, 1) \in S_2$, and every point $(x_1, x_2)$ of $\mathbf{R}^2$ is mapped to the point where the line joining the points $(0, 1, 1)$ and $(x_1, x_2, 0)$ in $\mathbf{R}^3$ meets $S_2$ again. This homeomorphism is known as stereographic projection.

Let $X$ be the union of a sequence $(K_n)$ of compact sets. Let $U_1$ be a relatively compact open neighbourhood of $K_1$ (no. 7, Proposition 10) and define $U_n$ inductively for $n > 1$ to be a relatively compact open neighbourhood of $\overline{U}_{n-1} \in K_n$ (no. 3, Proposition 5; no. 7, Proposition 10). The sequence $(U_n)$ clearly has the required properties.

For $K$ can be covered by a finite number of the $U_k$, by axiom (C''').

If $X$ is $\sigma$-compact we can construct a sequence of subsets $U_n$ of $X$ as in Proposition 15, and the neighbourhoods $X' - \overline{U}_n$ of $\omega$ in $X'$ form a fundamental system of neighbourhoods of $\omega$ by reason of Corollary 1. The converse follows from the fact that the complements of open neighbourhoods of $\omega$ are compact subsets of $X$.

Clearly every closed subspace of a locally compact $\sigma$-compact space is locally compact and $\sigma$-compact. Likewise any finite product of locally compact $\sigma$-compact spaces is locally compact and $\sigma$-compact.

Notice, however, that an open subspace of a compact space need not be $\sigma$-compact, as Alexandroff's theorem (no. 8, Theorem 4) shows.

Certainly $F$ is Hausdorff. On the other hand, if $(V_i)$ is an open covering in the subspace $F$, then each $V_i$ is of the form $V_i = U_i \cap F$, where $U_i$ is open in $X$. Consider the open covering $\mathscr{R}$ of $X$ formed by $CF$ and the $U_i$; since $X$ is paracompact, $\mathscr{R}$ has a locally finite refinement $\mathscr{R}'$, and the intersections with $F$ of the sets belonging to $\mathscr{R}'$ form a locally finite open covering of $F$ which is finer than the given covering $(V_i)$.

On the other hand an open subspace of a compact space need not be paracompact (Exercise 11).

Let $X$ be a paracompact space, $Y$ a compact space, $\mathscr{R}$ an open covering of $X \times Y$. For each $(x, y) \in X \times Y$ there is an open neighbourhood $V(x, y)$ of $x$ in $X$ and an open neighbourhood $W(x, y)$ of $y$ in $Y$ such that $V(x, y) \times W(x, y)$ is contained in some set belonging to $\mathscr{R}$. For each $x \in X$, the sets $W(x, y)$ as $y$ runs through $Y$ form an open covering of $Y$; hence there exist a finite number of points

such that the $W(x, y_i)$ cover $Y$. Let $U(x)$ denote then each of the open sets $U(x) \times W(x, y_i)$ is contained in a set of $\mathscr{R}$. Now let $(T_i)_{i \in I}$ be a locally finite open covering of $X$ which refines the covering $(U(x))_{x \in X}$. For each $i \in I$, let $x_i$ be a point of $X$ such that $T_i \subset U(x_i)$, and let us denote by $S_{i, k}$ the sets $W(x_i, y_k)$ corresponding to this point $x_i$ ($1 \leq k \leq n(x_i)$). Clearly the sets form an open covering of $X \times Y$ which refines $\mathscr{R}$, and the proof will be complete if we show that this covering is locally finite. Let $(x, y)$ be any point of $X \times Y$; there is a neighbourhood $Q$ of $x$ which meets only a finite number of sets $T_i$, and therefore the neighbourhood $Q \times Y$ of $(x, y)$ meets only a finite number of sets $T_i \times S_{i, k}$.

On the other hand the product of two paracompact spaces need not be paracompact (see Chapter IX, , Exercise 16)

Let $X$ be the sum of the $X_i$, and let $(V_\lambda)_{\lambda \in L}$ be an open covering of $X$. The covering formed by the open sets $X_i \cap Y_\lambda$ is finer than $(V_\lambda)$. For each $i \in I$, let $(U_{i,\mu})_{\mu \in M_i}$ be a locally finite open refinement of the covering $(V_\lambda \cap X_i)_{\lambda \in L}$; then the open covering of $X$ formed by the

is locally finite and refines the original covering $(V_\lambda)$.

Suppose $X$ is paracompact, and for each $x \in X$ let $V_x$ be a relatively compact open neighbourhood of $x$ in $X$. Then by hypothesis there exists a locally finite open covering $(U(\alpha))_{\alpha \in A}$ of $X$ which refines the covering $(V_x)_{x \in X}$. The $U(\alpha)$ are therefore relatively compact. Every compact subset $K$ of $X$ meets only a finite number of the sets $U(\alpha)$, for the non-empty sets $U(\alpha) \cap K$ form a locally finite open covering of the compact space $X$; hence they must be finite in number (no. 1). Now let $R$ be the following relation between two points $x, y$ of $X$: "there exists a finite sequence $(\alpha_i)_{1 \leq i \leq n}$ of indices in $A$ such that $x \in U(\alpha_1), \ y \in U(\alpha_n)$ and $U(\alpha_i)$ meets $U(\alpha_{i+1})$ for $1 \leq i \leq n - 1$".

It is immediately verified that $R$ is an equivalence relation, and that each equivalence class mod $R$ is an open subset of $X$ [since the $U(\alpha)$ are open sets]. $X$ is therefore the sum of the locally compact subspaces (no. 7, Proposition 13) formed by the equivalence classes mod $R$, and it remains to show that each of these subspaces is the union of a countable subfamily of the family $(U(\alpha))$.

Let $x$ be any point of $X$, and define a sequence $(C_n)$ of relatively compact open subsets of $X$ by induction on $n$ as follows; $C_1$ is the union of the sets $U(\alpha)$ which contain $x$, and for each $n > 1$, $C_n$ is the union of the sets $U(\alpha)$ which meet $C_{n-1}$. It is immediately verified by induction on $n$ that each of the $C_n$ is relatively compact and is the union of a finite number of sets $U(\alpha)$. Furthermore, the equivalence class of $x$ with respect to $R$ is the union of the $C_n$: for if $(\alpha_i)_{1 \leq i \leq n}$ is a sequence of indices such that $x \in U(\alpha_1)$ and $U(\alpha_i)$ meets $U(\alpha_{i+1})$ for $1 \leq i \leq n - 1$, then one sees by induction on $i$ that $U(\alpha_i) \subset C_i$ for $1 \leq i \leq n$. It follows that the equivalence classes mod $R$ are $\sigma$-compact, and this completes the proof of the first part of the theorem.

To prove the converse, we may assume (by Proposition 18) that $X$ is $\sigma$-compact. Let $\mathfrak{R} = (G_\lambda)_{\lambda \in L}$ be any open covering of $X$, and let $(U_n)$ be a sequence of relatively compact open sets in $X$ which have the properties stated in Proposition 15 of no. 9. Let $K_n$ denote the compact set $\overline{U}_n - U_{n-1}$ ($U_n = \emptyset$ if $n \leq 0$). The open set $U_{n+1} - \overline{U}_{n-2}$ is a neighbourhood of $K_n$ by construction; hence for each $x \in K_n$ there is a neighbourhood $W_x$ of $x$ contained in one of the sets $G_\lambda$ and contained also in $U_{n+1} - \overline{U}_{n-2}$. Since $K_n$ is compact, a finite number of the sets $W_x$ cover $K_n$; let $H_{ni} (1 \leq i \leq p_n)$ be these sets. Then the family $\mathfrak{R}'$ of sets $H_{ni} (n \geq 1, 1 \leq i \leq p_n \text{ for each } n)$ is an open covering of $X$ which refines $\mathfrak{R}$, and hence to complete the proof we have to show that $\mathfrak{R}'$ is locally finite. Let $z$ be any point of $X$, $n$ the smallest integer such that $z \in U_n$; then since $z \notin U_{n-1}$, there is a neighbourhood $T$ of $z$ which is contained in $U_n$ and does not meet $U_{n-2}$. It follows that $T$ meets only those sets $H_{mi}$ for which $n - 2 \leq m \leq n + 1$, i.e. $T$ meets only a finite number of sets of $\mathfrak{R}'$.

In the course of the proof we have also established the following result: