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
Every compact Hausdorff topological space is a normal topological space.
In fact a stronger statement hods: paracompact Hausdorff spaces are normal.
To prove this, consider the following lemma:
(separation by neighbourhoods of points from compact subspaces in Hausdorff spaces)
Let
$(X,\tau)$ be a Hausdorff topological space;
Then for every $x \in X \backslash Y$ there exists
an open neighbourhood $U_x \supset \{x\}$;
an open neighbourhood $U_Y \supset Y$
such that
By the assumption that $(X,\tau)$ is Hausdorff, we find for every point $y \in Y$ disjoint open neighbourhoods $U_{x,y} \supset \{x\}$ and $U_y \supset \{y\}$. By the nature of the subspace topology of $Y$, the restriction of all the $U_y$ to $Y$ is an open cover of $Y$:
Now by the assumption that $Y$ is compact, there exists a finite subcover, hence a finite set $S \subset Y$ such that
is still a cover.
But the finite intersection
of the corresponding open neighbourhoods of $x$ is still open, and by construction it is disjoint from all the $U_{s}$, hence in particular from their union
Therefore $U_x$ and $U_Y$ are two open subsets as required.
First we claim that $(X,\tau)$ is regular. To show this, we need to find for each point $x \in X$ and each disjoint closed subset $Y \in X$ dijoint open neighbourhoods $U_x \supset \{x\}$ and $U_Y \supset Y$. But since closed subspaces of compact spaces are compact, the subset $Y$ is in fact compact, and hence this is in fact the statement of lemma .
Next to show that $(X,\tau)$ is indeed normal, we apply the idea of the proof of lemma once more:
Let $Y_1, Y_2 \subset X$ be two disjoint closed subspaces. By the previous statement then for every point $y_1 \in Y$ we find disjoint open neighbourhoods $U_{y_1} \subset \{y_1\}$ and $U_{Y_2,y_1} \supset Y_2$. The union of the $U_{y_1}$ is a cover of $Y_1$, and by compactness of $Y_1$ there is a finite subset $S \subset Y$ such that
is an open neighbourhood of $Y_1$ and
is an open neighbourhood of $Y_2$, and both are disjoint.
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
maps from compact spaces to Hausdorff spaces are closed and proper
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
Last revised on April 28, 2017 at 07:36:08. See the history of this page for a list of all contributions to it.