topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
Let $X$ be a topological space which is
in the sense that for every open neighbourhood $U$ there exists a smaller open neighbourhood whose topological closure is compact and still contained in $U$;
Then there exists a countable open cover $\{U_i \subset X\}_{i \in \mathbb{N}}$ of $X$ such that for each $i \in I$
the topological closure $Cl(U_i)$ is a compact subspace
$Cl(U_i) \subset U_{i +1}$.
By sigma-compactness of $X$ there exists a countable cover $\{K_i \subset X\}_{i \in \mathbb{N}}$ of compact subspaces. We use these to construct the required cover by induction.
For $i = 0$ set
Then assume that for $n \in \mathbb{N}$ we have constructed a set $\{U_i \subset X\}_{i \in \{1, \cdots, n\}}$ with the required properties.
In particular this implies that
is a compact subspace. We now construct an open neighbourhood $U_{n+1}$ of this union as follows:
Let $\{U_x \subset X\}_{x \in Q_n}$ be a set of open neighbourhood around each of the points in $Q_n$. By local compactness of $X$, for each $x$ there is a smaller open neighbourhood $V_x$ with
So $\{V_x \subset X\}_{x \in Q_n}$ is still an open cover of $Q_n$. By compactness of $Q_n$, there exists a finite set $J_n \subset Q_n$ such that $\{V_x \subset X\}_{x \in J_n}$ is a finite open cover. The union
is an open neighbourhood of $Q_n$, hence in particular of $Cl(U_n)$. Moreover, since finite unions of compact spaces are compact (this prop.) and since the closure of a finite union is the union of the closures (this prop.), the closure of $U_{n+1}$ is compact:
This produces by induction a set $\{U_i \subset X\}_{i \in \mathbb{N}}$ with $Cl(U_i)$ compact and $Cl(U_i) \subset U_{i+1}$ for all $i \in \mathbb{N}$. It remains to see that this is a cover. This follows since by construction each $U_{i+1}$ is an open neighbourhood not just of $Cl(U_{i})$ but in fact of $Q_i$, hence in particular of $K_i$, and since the $K_i$ form a cover:
(locally compact and sigma-compact spaces are paracompact)
Let $X$ be a topological space which is
Then $X$ is also paracompact.
Let $\{U_i \subset X\}_{i \in I}$ be an open cover of $X$. We need to show that this has a refinement by a locally finite cover.
By lemma there exists a countable open cover $\{V_n \subset X\}_{n \in \mathbb{N}}$ of $X$ such that for all $n \in \mathbb{N}$
$Cl(V_n)$ is compact;
$Cl(V_n) \subset V_{n+1}$.
Notice that the complement $Cl(V_{n+1}) \setminus V_n$ is compact, since $Cl(V_{n+1})$ is compact and $V_n$ is open (by this lemma).
By this compactness, the cover $\{U_i \subset X\}_{i \in I}$ regarded as a cover of the subspace $Cl(V_{n+1})\setminus V_n$ has a finite subcover $\{U_i \subset X\}_{i \in J_n}$ indexed by a finite set $J_n \subset I$, for each $n \in \mathbb{N}$.
We consider the sets of intersections
Since $V_{n+2} \setminus Cl(V_{n-1})$ is open, and since $Cl(V_{n+1}) \subset V_{n+2}$ by construction, this is still an open cover of $Cl(V_{n+1})\setminus V_n$. We claim now that
is a locally finite refinement of the original cover, as required:
$\mathcal{U}$ is a refinement, since by construction each element in $\mathcal{U}_n$ is contained in one of the $U_i$;
$\mathcal{U}$ is still a covering because by construction it covers $Cl(V_{n+1}) \setminus V_n$ for all $n \in \mathbb{N}$, and since by the nested nature of the cover $\{V_n \subset X\}_{n \in \mathbb{N}}$ also $\{Cl(V_{n+1}) \setminus V_n\}_{n \in \mathbb{N}}$ is a cover of $X$.
$\mathcal{U}$ is locally finite because each point $x \in X$ has an open neighbourhood of the form $V_{n+2} \setminus Cl(V_{n-1})$ (since these also form an open cover, by the nestedness) and since by construction this has trivial intersection with $\mathcal{U}_{\geq n+3}$ and since all $\mathcal{U}_n$ are finite, so that also $\underset{k \lt n+3}{\cup} \mathcal{U}_k$ is finite.
Last revised on October 27, 2018 at 22:15:45. See the history of this page for a list of all contributions to it.