topology (point-set topology, point-free topology)
see also 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
A topological space is called locally path-connected if it has a basis of path-connected neighbourhoods. In other words, if for every point $x$ and neighbourhood $V \ni x$, there exists a path-connected neighbourhood $U \subset V$ that contains $x$.
Let $X$ be a locally path connected space. Then the path connected component $P_x \subset X$ over any point $x \in X$ is an open set.
It is sufficient to show that every point $y \in P_x$ has an neighbourhood $U_y$ which is still contained in $P_x$. But by local path connectedness, $y$ has a neighbourhood $V_y$ which is path connected. It follows by concatenation of paths that $V_y \subset P_x$.
A locally path-connected space is connected if and only if it is path-connected.
The connected components of a locally path-connected space are the same as its path-connected components.
A path connected component is always connected (this lemma), and in a locally path-connected space is it also open (lemma 1). This means that every path-connected component is also connected.
Conversely, it is now sufficient to see that every connected component is path-connected. Suppose it were not, then it would be covered by more than one disjoint non-empty path-connected components. But by lemma 1 these would be all open. This would be in contradiction with the assumption that $U$ is connected. Hence we have a proof by contradiction.
(Euclidean space is locally path-connected)
For $n \in \mathbb{N}$ then Euclidean space $\mathbb{R}^n$ (with its metric topology) is locally path-connected, since each open ball is path-connected topological space.
Every open subspace of a locally path connected topological space is itself locally path connected.
The condition is a necessary assumption in the
in the form of the condition for