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
For any $a \lt b \in \mathbb{R}$ the closed interval
regarded with its subspace topology of Euclidean space with its metric topology is a compact topological space.
Since all the closed intervals are homeomorphic it is sufficient to show the statement for $[0,1]$. Hence let $\{U_i \subset [0,1]\}_{i \in I}$ be an open cover. We need to show that it has an open subcover.
Say that an element $x \in [0,1]$ is admissible if the closed sub-interval $[0,x]$ is covered by finitely many of the $U_i$. In this terminology, what we need to show is that $1$ is admissible.
Observe from the definition that
0 is admissible,
if $y \lt x \in [0,1]$ and $x$ is admissible, then also $y$ is admissible.
This means that the set of admissible $x$ forms either
an open interval $[0,g)$
or a closed interval $[0,g]$,
for some $g \in [0,1]$. We need to show that the latter is true, and for $g = 1$. We do so by observing that the alternatives lead to contradictions:
Assume that the set of admissible values were an open interval $[0,g)$. Pick an $i_0 \in I$ such that $g \in U_{i_0}$ (this exists because of the covering property). Since such $U_{i_0}$ is an open neighbourhood of $g$, there is a positive real number $\epsilon$ such that the open ball $B^\circ_g(\epsilon) \subset U_{i_0}$ is still contained in the patch. It follows that there is an element $x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g)$ and such that there is a finite subset $J \subset I$ with $\{U_i \subset [0,1]\}_{i \in J \subset I}$ a finite open cover of $[0,x)$. It follows that $\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\}$ were a finite open cover of $[0,g]$, hence that $g$ itself were still admissible, in contradiction to the assumption.
Assume that the set of admissible values were a closed interval $[0,g]$ for $g \lt 1$. By assumption there would then be a finite set $J \subset I$ such that $\{U_i \subset [0,1]\}_{i \in J \subset I}$ were a finite cover of $[0,g]$. Hence there would be an index $i_g \in J$ such that $g \in U_{i_g}$. But then by the nature of open subsets in the Euclidean space $\mathbb{R}$, this $U_{i_g}$ would also contain an open ball $B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon)$. This would mean that the set of admissible values includes the open interval $[0,g+ \epsilon)$, contradicting the assumption.
This gives a proof by contradiction.
Last revised on June 6, 2017 at 04:20:03. See the history of this page for a list of all contributions to it.