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
(Dieudonné‘s theorem)
Every paracompact Hausdorff space is normal.
In particular compact Hausdorff spaces are normal.
Let $(X,\tau)$ be a paracompact Hausdorff space
We first show that it is regular: To that end, let $x \in X$ be a point, and let $C \subset X$ be a closed subset not containing $x$. We need to find disjoint open neighbourhoods $U_x \supset \{x\}$ and $U_C \supset C$.
First of all, by the Hausdorff property there exists for each $c \in C$ disjoint open neighbourhods $U_{x,c} \supset \{x\}$ and $U_c \supset \{c\}$. As $c$ ranges, the latter clearly form an open cover $\{U_c \subset X\}_{c \in C}$ of $C$, and so the union
is an open cover of $X$. By paracompactness of $(X,\tau)$, there exists a locally finite refinement, and by this lemma we may assume its elements to share the original index set and be contained in the original elements of the same index. Hence
is a locally finite collection of subsets, such that
is an open neighbourhood of $C$.
Now by definition of local finiteness there exists an open neighbourhood $W_x \supset \{x\}$ and a finite subset $K \subset C$ such that
Consider then
which is an open neighbourhood of $x$, by the finiteness of $K$.
It thus only remains to see that
But this holds because the only $V_{c}$ that intersect $W_x$ are the $V_{k} \subset U_{k}$ for $k \in K$ and each of these is by construction disjoint from $U_{x,k}$ and hence from $U_x$.
This establishes that $(X,\tau)$ is regular. Now we prove that it is normal. For this we use the same approach as before:
Let $C,D \subset X$ be two disjoint closed subsets. We need to produce disjoint open neighbourhoods for these.
By the previous statement of regularity, we may find for each $c \in C$ disjoint open neighbourhoods $U_c \supset \{c\}$ and $U_{D,c} \supset D$. Hence the union
is an open cover of $X$, and thus by paracompactness has a locally finite refinement, whose elements we may, again by this lemma, assume to have the same index set as before and be contained in the previous elements with the same index. Hence we obtain a locally finite collection of subsets
such that
is an open neighbourhood of $C$.
It is now sufficient to see that every point $d \in D$ has an open neighbourhood $U_d$ not intersecting $U_C$, for then
is the required open neighbourhood of $D$ not intersecting $U_C$.
Now by local finiteness of $\{V_c \subset X\}_{c \in C}$, every $d \in D$ has an open neighbourhood $W_d$ such that there is a finite set $K_d \subset C$ so that
Accordingly the intersection
is still open and disjoint from the remaining $V_k$, hence disjoint from all of $U_C$.
Last revised on June 24, 2017 at 20:28:10. See the history of this page for a list of all contributions to it.