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
Given a topological space (or locale) $X$, a subspace $A$ of $X$ is dense if its closure is all of $X$: $cl(A)=X$.
Since $cl(A)$ is the set of all points $x$ such that every open neighborhood of $x$ intersects $A$, this can equivalently be written as “every open neighborhood of every point intersects $A$”, or equivalently “every inhabited open set intersects $A$”, i.e. $A\cap U$ is inhabited for all inhabited open sets $U$.
Contraposing this, we obtain another equivalent definition “the only open subset not intersecting $A$ is the empty set”, or “if $A\cap U=\emptyset$ for some open set $U$, then $U=\emptyset$”. This is the definition usually given when $X$ is a locale: a nucleus $j$ is dense if $j(0)=0$ (since $j(0)$ is the union of all opens whose “intersection with $j$” is $0$).
If $A \subseteq X$ is a dense subset of a topological space $X$ and $f: X \to Y$ is an epimorphism, then the image $f(A)$ is dense in $Y$.
If i: $A \hookrightarrow X$ and $j: X \hookrightarrow B$ are dense subspace inclusions, then so is the composite $j \circ i: A \to B$.
If $A\subseteq X$ is a dense subset of topological space $X$ and $A$ is connected, so is $X$.
In point-set topology, a space is separable if and only if it has a dense subspace with countably many points.
In locale theory, we have the curious property that any intersection of dense subspaces is still dense. (This of course fails rather badly for topological spaces, where the intersection of all dense topological subspaces is the space of isolated points.) One consequence is that every locale has a smallest dense sublocale, the double negation sublocale.
In the category of Hausdorff topological spaces (with continuous functions between them), the inclusion of a dense subspace
is an epimorphism.
We have to show that for $(f,g)$ any pair of parallel morphisms out of $X$
into a Hausdorff space $Y$, the equality $f \circ i = g \circ i$ implies $f = g$. With classical logic we may equivalently show the contrapositive: That $f \neq g$ implies $f \circ i \neq g \circ i$.
So assume that $f \neq g$. This means that there exists $y \in Y$ with $f(y) \neq g(y)$. But since $Y$ is Hausdorff, there exist disjoint open neighbourhoods $O_{f(y)},\;O_{g(y)} \subset Y$, i.e. $f(x) \in O_{f(x)}$ and $g(x) \in O_{g(x)}$ with $O_{f(x)} \cap O_{g(x)} = \varnothing$.
But their preimages must intersect at least in $x \in f^{-1}\big( O_{f(x)} \big) \cap g^{-1}\big( O_{g(x)} \big)$. Since this intersection is an open subset (as preimages of open subsets are open by definition of continuous functions, and since finite intersections of open subsets are open by the definition of topological spaces) there exists a point $a \in A$ with $i(a) \in f^{-1}\big( O_{f(x)} \big) \cap g^{-1}\big( O_{g(x)} \big)$ (by definition of dense subset). But since then $f(i(a)) \in O_{f(x)}$ and $g(i(a)) \in O_{g(x)}$ while $O_{f(x)}$ is disjoint from $O_{g(x)}$, it follows that $f(i(a)) \neq g(i(a))$. This means that $f \circ i \neq g \circ i$.
In constructive mathematics, the law of contraposition is not an equivalence, so we obtain two inequivalent notions of density:
Of course, strong density implies weak density, since emptiness is non-inhabitation (whereas inhabitation is stronger than non-emptiness). The two notions of density are related dually to the corresponding notions of closed subspace: $A$ is strongly dense iff its weak closure is all of $X$, and weakly dense iff its strong closure is all of $X$.
Note that the usual notion of density for sublocales $j(0)=0$ is an analogue of weak density, and could be called such. There is also a notion of strong density for sublocales. Since strong density refers to inhabited sets, one might expect strong density for sublocales to refer to positive elements, and thus only be sensible for overt locales; but in fact it can be reformulated to make sense in all cases.
A nucleus $j$ on a locale $X$ is strongly dense if $j(\hat{P})=\hat{P}$ for any truth value $P$, where $\hat{P} = \bigvee \{ X \mid P \}$.
With classical logic, every truth value is either $\top$ or $\bot$, and we have $\hat{\top}=X$ (and any nucleus satisfies $j(X)=X$) while $\hat{\bot}=0$. Thus classically strong and weak density coincide. To see that this is really a notion of strong density, we prove:
If $i:A\subseteq X$ is a sublocale such that $A$ and $X$ are both overt, then $A$ is strongly dense if and only if for any positive open $U\in O(X)$, the intersection $A\cap U = i^*U \in O(A)$ is also positive.
First suppose $A$ is strongly dense, and let $U\in O(X)$ be positive. Let $P$ be the truth value of the statement “$i^*U \in O(A)$ is positive”. We want to show that $P$ is true, for which it suffices to show that $\hat{P} = \bigvee \{ X \mid P \}$ is positive, since then its covering $\{ X \mid P \}$ would be inhabited and thus $P$ would be true. And since $U$ is positive, it suffices to show $U \subseteq \hat{P}$.
Now since $A$ is strongly dense, $j_A(\hat{P}) = \hat{P}$, which is to say that $\hat{P} = i_*(i^*\hat{P})$. By adjointness, therefore, to show $U \subseteq \hat{P}$ it suffices to show $i^*U \subseteq i^*\hat{P} = \bigvee \{ A \mid P\}$. Now since $A$ is overt, $i^*U$ can be covered by positive opens, so it suffices to show that for any positive $V\subseteq i^*U$ we have $V\subseteq \bigvee \{ A \mid P\}$. But if $V\subseteq i^*U$ is positive, then $i^*U$ is also positive, i.e. $P$ is true, and thus $\bigvee \{ A \mid P\} = A$, which contains $V$.
Now suppose conversely that for any positive $U\in O(X)$, $i^*U$ is also positive, and let $P$ be any truth value; we must show $i_* i^*\hat{P} \subseteq \hat{P}$. Since $X$ is overt, $i_* i^*\hat{P}$ can be covered by positive opens, so it suffices to show that for any positive $U\subseteq i_* i^*\hat{P}$ we have $U\subseteq \hat{P}$. But by adjointness, $U\subseteq i_* i^*\hat{P}$ is equivalent to $i^*U \subseteq i^*\hat{P}$, and by assumption $i^*U$ is also positive. Thus, $i^*\hat{P} = \bigvee \{ A \mid P\}$ is positive, which means that $P$ is true, and hence $\hat{P} = X$ and so $U\subseteq \hat{P}$.
Since spatial locales are overt, and their positivity predicate coincides with inhabitedness, we have in particular:
If $i:A\subseteq X$ is a subspace of a topological space, then $A$ is strongly dense as a topological subspace if and only if it is strongly dense as a sublocale.
Strong density for sublocales gives rise to a corresponding notion of weakly closed sublocale. It is also the specialization of the notion of fiberwise dense sublocale? to the case of locale maps $X\to 1$.
Strongly dense sublocales are discussed in
Sketches of an Elephant, C1.1 and C1.2
Peter Johnstone, A constructive ‘closed subgroup theorem’ for localic groups and groupoids, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1989), Volume: 30, Issue: 1, page 3-23 link
Mamuka Jibladze, Peter Johnstone, The frame of fibrewise closed nuclei, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1991), Volume: 32, Issue: 2, page 99-112, link
Peter Johnstone, Fiberwise separation axioms for locales
Last revised on May 31, 2022 at 15:57:09. See the history of this page for a list of all contributions to it.