Topology: General Topology 2021

Suppose \(f\) closed, and suppose \(Z\subseteq X\) a closed subset; then

but the other inclusion \(f(Z)\subseteq\overline{f(Z)}\) is automatic. Hence \(f(Z)=\overline{f(Z)}\text{.}\)

Conversely, suppose that \(f\) satisfies the condition that for any closed subset \(Z\subseteq X\text{,}\) the direct image \(f(Z)\) is closed, and suppose \(S\subseteq X\) any subset. Then \(f(\overline{S})\) is certainly closed, and of course it contains \(f(S)\text{;}\) hence it contains \(\overline{f(S)}\text{.}\)

Let \(Z \subseteq Y\) be a subset such that \(f^{-1}(Z) \subseteq X\) is closed. Then \(Z = f(f^{-1}(Z)) \subseteq Y\) is closed, since \(f\) is closed.

Assume that each restriction \(f|_{W_j}\) is closed, and let \(Z\subseteq X\) a closed subset. Then \(Z\cap W_j\) is closed in \(W_j\text{,}\) and so the subset

is closed, whence so is the subset \(f(Z)=\bigcup_{j=1}^nf(Z\cap W_j)\text{.}\)

Assume that \(f\) is closed, and assume that \(y\in Y\) and \(U\supseteq X_y\) open. Then set \(V\coloneq Y \smallsetminus f(X \smallsetminus U)\text{;}\) clearly \(y\in V\text{,}\) and since \(f\) is closed, \(V\) is open. Now an element of \(f^{-1}(V)\) does not lie in \(f^{-1}(f(X \smallsetminus U))\text{,}\) whence we deduce that it does not lie in \(X \smallsetminus U\text{,}\) whence it lies \(U\text{,}\) as desired.

Conversely, suppose the second condition is satisfied, and suppose \(Z\subseteq X\) a closed subset. By the second condition, for any point \(y\notin f(Z)\text{,}\) there is an open neighbourhood \(V\) of \(y\) such that \(f^{-1}(V)\subseteq X \smallsetminus Z\text{,}\) whence we deduce that \(Y \smallsetminus f(Z)\) is open.

Let us assume the first condition and prove the second. So let \(T\) be a topological space and \(t\in T\text{,}\) and let \(U\) an open set of \(T\times X\) containing \(\{t\}\times X\text{.}\) Now let \(\UU\) be the collection of those open subsets \(W\subseteq X\) such that for some open neighbourhood \(E\) of \(t\text{,}\) one has \(E\times W\subseteq U\text{.}\) The collection \(\UU\) covers

\(X\text{.}\) The first condition now ensures that there is a finite subcover \(\UU_0\subseteq\UU\text{.}\) For each \(W\in\UU_0\text{,}\) select an open neighbourhood \(E_W\) of \(t\) such that \(E_W\times W\subseteq U\text{.}\) Now set

so that \(V\) is an open neighbourhood of \(t\text{,}\) and \(V\times W\subseteq U\) for any \(W\in\UU_0\text{.}\) Hence

This verifies the second condition.

The equivalence of the second and third conditions is the content of the previous proposition.

Let us assume the third condition and prove the fourth. So let \(\ZZ\subseteq\PP(X)\) be a collection of closed subsets of \(X\text{,}\) and assume that for any finite subset \(\ZZ_0\subseteq\ZZ\text{,}\) the intersection \(\cap\ZZ_0\) is nonempty. Now let \(T\) be the set \(X\sqcup\{\infty\}\text{,}\) which we topologise with the coarsest topology such that

• any subset \(S\subseteq X\) is open in \(T\text{,}\) and

• for any \(Z\in\ZZ\text{,}\) the set \(Z\sqcup\{\infty\}\) is open in \(T\text{.}\)

Any open neighbourhood of \(\infty\) contains a finite intersection of elements of \(\ZZ\text{.}\) Since all such intersections are nonempty, any open neighbourhood of \(\infty\) contains at least one point of \(X\text{.}\) Hence \(\infty\) is not open or, equivalently, \(X\subseteq T\) is dense. Now let

and consider the closed set \(\pr_1(\overline{\Delta})\subseteq T\text{.}\) Clearly \(X=\pr_1(\Delta)\subseteq\pr_1(\overline{\Delta})\text{,}\) so \(\pr_1(\overline{\Delta})=T\text{.}\) In other words, \(\overline{\Delta}\) contains a point \((\infty, x)\) for some \(x\in X\text{.}\) Thus for every \(Z\in\ZZ\) and every open neighbourhood \(U\) of \(x\text{,}\) one has \(((Z\sqcup\{\infty\})\times U)\cap\Delta\neq\varnothing\text{,}\) and so

\(Z\cap U\neq\varnothing\text{.}\) Thus \(x\) is close to every \(Z\in\ZZ\text{,}\) whence \(x\) lies in every \(Z\in\ZZ\text{,}\) whence \(x\in\cap\ZZ\text{.}\)

Finally, let's assume the fourth condition and prove the first. Let \(\UU\subseteq\PP(X)\) be an open cover of \(X\text{.}\) Now let \(\ZZ\) be the set of complements of elements

of \(\UU\text{.}\) Clearly \(\cap\ZZ=X-\cup\UU=\varnothing\text{,}\) so by the fourth condition, there exists a finite subset \(\ZZ_0\subseteq\ZZ\) such that \(\cap\ZZ_0=\varnothing\text{.}\) Now let \(\UU_0\) be the set of complements of elements

of \(\ZZ_0\text{.}\) Now \(\cup\UU_0=X \smallsetminus \cap\ZZ_0=X\text{,}\) whence \(\UU_0\subseteq\UU\) is a finite subcover.

Suppose \(\UU\) an open covering of \([a,b]\text{;}\) then let

Now suppose, to generate a contradiction, that \(c\lt b\text{;}\) then let \(U\in\UU\) be an element of the open cover containing \(c\text{.}\) Then for some \(\varepsilon>0\text{,}\) one has \(\left]c-\varepsilon,c+\varepsilon\right[\subset U\text{.}\) By definition of \(c\text{,}\) one has \([a,c-\varepsilon/2]\subseteq\cup\UU_0\) for a finite subset \(\UU_0\subseteq\UU\text{;}\) hence the union of the elements of \(\UU_0\cup\{U\}\subseteq\UU\) contains \([a,c+\varepsilon/2]\text{,}\) yielding the desired contradiction. Thus \(c=b\text{.}\)

Now choose an element \(V\in\UU\) than contains \(b\text{;}\) for some \(\varepsilon>0\text{,}\) one has \(\left]b-\varepsilon,b\right]\subset U\text{.}\) Now since \(c=b\text{,}\) there exists a finite subset \(\VV_0\subseteq\UU\) such that \(\left[a,b-\varepsilon/2\right[\subseteq\cup\VV_0\text{,}\) and so \(\VV_0\cup\{V\}\subseteq\UU\) is a finite cover of \([a,b]\text{.}\)

Let \(T\) be a topological space. If \(Z\subseteq T\times Y\) is closed, then \(f^{-1}(Z)\subseteq X\) is closed, so

is closed as well.

Let \(X\) be a compactum, and suppose \(Z\subseteq X\) closed. For any topological space \(T\text{,}\) the projection \(\pr_1\colon\fromto{T\times X}{T}\) is closed, and the inclusion \(T\times Z \inclusion T\times X\) is closed, so the composite \(\fromto{T\times Z}{T}\text{,}\) which is the projection, is closed.

Suppose \(T\) a topological space. Then the projections \(\fromto{T\times X_1}{T}\text{,}\) \(\fromto{T\times X_1\times X_2}{T\times X_1}\text{,}\) …, and \(\fromto{T\times X_1\times X_2\times\cdots\times X_n}{T\times X_1\times\cdots\times X_{n-1}}\) are all closed, and so their composite

is closed as well.

Let us show that the first of these conditions implies the second. Suppose \(A\) compact, and suppose \((x_i)_{i\geq 0}\) a sequence of points in \(A\text{.}\) If there were no convergent subsequence of \((x_i)_{i\geq 0}\text{,}\) then the set \(\{x_i\}_{i\geq 0}\) would be a closed subset of \(A\text{,}\) hence compact, and there would be a sequence \(\varepsilon_i\) of positive real numbers such that the balls \(B^n(x_i,\varepsilon_i)\) would be disjoint. But then \(\{B^n(x_i,\varepsilon_i)\}_{i\geq 0}\) would be an open cover of \(\{x_i\}_{i\geq 0}\) with no finite subcover.

Let us show that the second property implies the third. Suppose \(A\) has the property that every sequence has a convergent subsequence. Then \(A\) must be bounded, since otherwise there exists a sequence \((x_i)_{i\geq 0}\) of points in \(A\) such that \(\|x_i\|\to\infty\text{.}\) It must also be closed, since if \(x\in\overline{A}-A\text{,}\) one can construct a sequence of points of \(A\) converging to \(x\text{.}\)

Finally, let us show that the third condition implies the first. Suppose \(A\) is closed and bounded. Since \(A\) is bounded, it is contained in a box \([-a,a]^n\) for some \(a\geq 0\text{.}\) Since every closed subspace of a compact space is compact, it is enough to show that \([-a,a]^n\) is compact. Since a finite product of compacta is compact, it is enough to show that any closed interval is compact, and we've already shown this.