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
It is not generally true that a topological space is the disjoint union space (coproduct in Top) of its connected components. The spaces such that this is true for all open subspaces are the locally connected topological spaces.
(locally connected topological space)
A topological space $X$ is locally connected if every point has a neighborhood basis of connected open subsets.
(alternative characterizations of local connectivity)
For $X$ a topological space, then the following are equivalent:
every connected component of every open subspace of $X$ is open;
every open subset, as a topological subspace, is the disjoint union space (coproduct in Top) of its connected components.
In particular, in a locally connected space, every connected component $S$ is a clopen subset; hence connected components and quasi-components coincide.
$\,$
1) $\Rightarrow$ 2)
Assume $X$ is locally connected, and let $U \subset X$ be an open subset with $U_0 \subset U$ a connected component. We need to show that $U_0$ is open.
Consider any point $x \in U_0$. Since then also $x \in U$, the defintion of local connectedness, def. , implies that there is a connected open neighbourhood $U_{x,0}$ of $X$. Observe that this must be contained in $U_0$, for if it were not then $U_0 \cup U_{x,0}$ were a larger open connected open neighbourhood, contradicting the maximality of the connected component $U_0$.
Hence $U_0 = \underset{x \in U_0}{\cup} U_{x,0}$ is a union of open subsets, and hence itself open.
2) $\Rightarrow$ 3)
Now assume that every connected component of every open subset $U$ is open. Since the connected components generally consitute a cover of $X$ by disjoint subsets this means that now they for an open cover by disjoint subsets. But by forming intersections with the cover this implies that every open subset of $U$ is the disjoint union of open subsets of the connected components (and of course every union of open subsets of the connected components is still open in $U$), which is the definition of the topology on the disjoint union space of the connected components.
3) $\Rightarrow$ 1)
Finally assume that every open subspace is the disjoint union of its connected components. Let $x$ be a point and $U_x \supset \{x\}$ a neighbourhood. We need to show that $U_x$ contains a connected neighbourhood of $x$.
But, by definition, $U_x$ contains an open neighbourhood of $x$ and by assumption this decomposes as the disjoint union of its connected components. One of these contains $x$. Since in a disjoint union space all summands are open, this is the required connected open neighbourhod.
(Euclidean space is locally connected)
For $n \in \mathbb{N}$ the Euclidean space $\mathbb{R}^n$ (with its metric topology) is locally connected.
By nature of the Euclidean metric topology, every neighbourhood $U_x$ of a point $x$ contains an open ball containing $x$. Moreover, every open ball clearly contains an open cube, hence a product space $\underset{i \in \{1, \cdots, n\}}{\prod} (x_i-\epsilon, x_i + \epsilon)$ of open intervals which is still a neighbourhood of $x$.
Now intervals are connected (by this example) and products of connected spaces are connected (by this example). This shows that ever open neighbourhood contains a connected neighbourhood.
(open subspace of locally connected space is locally connected)
Every open subspace of a locally connected space is itself locally connected
The topologist's sine curve is connected but not locally connected.
Examples of locally connected spaces include topological manifolds.
Finally,
A space $X$ is totally disconnected topological space if its connected components are precisely the singletons of $X$.
In other words, a space is totally disconnected if its coreflection into $LocConn$ is discrete. Such spaces recur in the study of Stone spaces.
The category of totally disconnected spaces is a reflective subcategory of $Top$. The reflector sends a space $X$ to the space $X/\sim$ whose points are the connected components of $X$, endowed with the quotient topology induced by the projection $q: X \to X/\sim$. Details may be found at totally disconnected space.
Let $i \colon LocConn \hookrightarrow Top$ be the full subcategory inclusion of locally connected spaces into all of Top. The following result is straightforward but useful.
$LocConn$ is a coreflective subcategory of $Top$, i.e., the inclusion $i$ has a right adjoint $R$. For $X$ a given space, $R(X)$ has the same underlying set as $X$, topologized by letting connected components of open subspaces of $X$ generate a topology.
Being a coreflective category of a complete and cocomplete category, the category $LocConn$ is also complete and cocomplete. Of course, limits and particularly infinite products in $LocConn$ are not calculated as they are in $Top$; rather one takes the limit in $Top$ and then retopologizes it according to Theorem . (For finite products of locally connected spaces, we can just take the product in $Top$ – the result will be again locally connected.)
Let $\Gamma \colon LocConn \to Set$ be the underlying set functor, and let $\nabla, \Delta \colon Set \to LocConn$ be the functors which assign to a set the same set equipped with the codiscrete and discrete topologies, respectively. Let $\Pi_0 \colon LocConn \to Set$ be the functor which assigns to a locally connected space the set of its connected components.
There is an adjoint quadruple of adjoint functors
and moreover, the functor $\Pi_0$ preserves finite products.
The proof is largely straightforward; we point out that the continuity of the unit $X \to \Delta \Pi_0 X$ is immediate from a locally connected space’s being the coproduct of its connected components. As for $\Pi_0$ preserving finite products, write locally connected spaces $X$, $Y$ as coproducts of connected spaces
then their product in $LocConn$ coincides with their product in $Top$, and is
where each summand $C_i \times D_j$ is connected by Result . From this it is immediate that $\Pi_0$ preserves finite products.
Accordingly the category of sheaves on a locally connected space is a locally connected topos. For related discussions, see also cohesive topos.
A quotient space of a locally connected space $X$ is also locally connected.
Suppose $q: X \to Y$ is a quotient map, and let $V \subseteq Y$ be an open neighborhood of $y \in Y$. Let $C(y)$ be the connected component of $y$ in $V$; we must show $C(y)$ is open in $Y$. For that it suffices that $C = q^{-1}(C(y))$ be open in $X$, or that each $x \in C$ is an interior point. Since $X$ is locally connected, the connected component $U_x$ of $x$ in $q^{-1}(V)$ is open, and the subset $q(U_x) \subseteq V$ is connected, and therefore $q(U_x) \subseteq C(y)$ (as $C(y)$ is the maximal connected subset of $V$ containing $q(x)$). Hence $U_x \subseteq q^{-1}(C(y)) = C$, proving that $x$ is interior to $C$, as desired.
The conclusion does not follow if $q: X \to Y$ is merely surjective; e.g., there is a surjective (continuous) map from $\mathbb{R}$ to (a version of) the Warsaw circle, but the latter is not locally connected.
Last revised on June 12, 2017 at 15:29:15. See the history of this page for a list of all contributions to it.