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 two topological spaces $(X_1,\tau_1)$ and $(X_2, \tau_2)$, then the Cartesian product of their underlying sets $X_1 \times X_2$ is naturally equipped with a topology $\tau_{X_1 \times X_2}$ itself, generated from the base opens which are themselves Cartesian product $U_1 \times U_2 \subset X_1 \times X_2$, of open subsets of the original spaces: $U_i \subset X_i$. The resulting topological space
is called the product topological space of the two original spaces.
The formally dual concept is that of disjoint union topological spaces.
More generally, consider any index set $I$ and an $I$-indexed set $\{X_i, \tau_i\}_{i \in I}$ of topological spaces. Then the product topology $\tau_{prod}$ or Tychonoff topology on the Cartesian product $\underset{i \in I}{\prod} X_i$ of underlying sets is equivalently
the topology generated from the sub-base given by products $\underset{i \in I}{\prod} U_i$ with $U_i \subset X_i$ open, but all except one of the factors equal to the corresponding $X_i$,
hence the topology whose open subsets are precisely those obtained as arbitrary unions of finite intersections of such subsets;
the topology generated from the base given by products $\underset{i \in I}{\prod} U_i$ with $U_i \subset X_i$ open, but all except a finite number of factors equal to the corresponding $X_i$,
hence the topology whose open subsets are precisely those obtained as arbitrary unions of such subsets.
This product topology is singled out by the fact that the resulting product topological space is the category theoretic product of the original space in the category Top of topological spaces:
This means that the product topology enjoys the universal property that for any topological space $(Y,\tau_Y)$ then sets of continuous functions $\{(Y, \tau_Y) \overset{\phi_i}{\to} (X_i, \tau_i)\}_{i \in I}$ into the factor spaces are in natural bijection with continuous functions $(\phi_i)_{i \in I} \colon (Y, \tau_Y) \to \left(\underset{i \in I}{\prod} X_i, \tau_{prod}\right)$ into the product topological space.
Beware that (among others) there is also the box topology $\tau_{box}$ on the Cartesian product of underlying sets $\underset{i\in I}{\prod} X_i$, whose open subsets are the unions of those of the for $\underset{i \in I}{\prod} U_i$ with $U_i \subset X_i$ open and with no further restriction on the factors.
For $I$ a finite set, then these two topologies coincide, but for $I$ not finite then the box topology is a strictly finer topology
and hence in this case it does not enjoy the universal property of the product topology above.
examples of universal constructions of topological spaces:
$\phantom{AAAA}$limits | $\phantom{AAAA}$colimits |
---|---|
$\,$ point space$\,$ | $\,$ empty space $\,$ |
$\,$ product topological space $\,$ | $\,$ disjoint union topological space $\,$ |
$\,$ topological subspace $\,$ | $\,$ quotient topological space $\,$ |
$\,$ fiber space $\,$ | $\,$ space attachment $\,$ |
$\,$ mapping cocylinder, mapping cocone $\,$ | $\,$ mapping cylinder, mapping cone, mapping telescope $\,$ |
$\,$ cell complex, CW-complex $\,$ |
Let $\{(X_i, \tau_i)\}_{i \in I}$ be a set of topological spaces. Then their product topological space has as underlying set the Cartesian product $\underset{i \in I}{\prod} X_i$ of the underlying sets, and has as topology $\tau_{prod} \subset P(\underset{i \in I}{\prod} X_i)$ the coarsest topology such that all the projection maps
become continuous functions (calle the Tychonoff topology).
This means equivalently that $\tau_{prod}$ is the topology generated from the sub-base
Notice that
and that
etc.
Let $X$ be any topological space and write $Disc(\{1,2\})$ for the discrete topological space on a set with two elements. Then there is a homeomorphism
between the product space of $X$ with $Disc(\{1,2\})$ and the disjoint union space of $X$ with itself.
By definition of disjoint union there is a bijection of underlying sets $X \sqcup X \simeq X \times \{1,2\}$.
By unwinding the definitions
the open subsets of $X \times Disc(\{0,1\})$ are unions of those of the form $U \times S$, where $U \subset X$ is an open subset and $S \subset \{1,2\}$ is any subset
the open subsets of $X \sqcup X$ are of the form $U \sqcup V$ with $U,V \subset X$ open.
Under the above bijection the we have
Conversely, every union of elements of the form $U' \times \{1\}$, $V' \times \{2\}$ and $W \times \{1,2\} = W \times \{1\} \cup W \times \{2\}$ is of the form $U \times \{1\} \cup V \times \{2\}$.
This shows that the above bijection of underlying sets induces a bijection of the corresponding open subsets, hence is a homeomorphism.
For $n \in \mathbb{N}$ consider the Cartesian space $\mathbb{R}^n$ with the metric topology induced from its Euclidean metric structure. Then the product topological spaces satisfy
Let $Disc(S)$ be a discrete topological space on a set with at least two elements. Then the infinite product space
is itself not a discrete space.
The open subsets of a discrete space include all the subsets of the underlying set. Hence we need to see that there are subsets of the Cartesian product set $\underset{n \in \mathbb{Z}}{\prod} Disc(S)$ which are not open subsets in the Tychonoff topology.
But by definition the open subsets in the Tychnoff topology are unions of products of open subsets of the factor spaces such that only a finite number of the factors is not the total space.
Since $S$ is assumed to contain at least two elements let $1 \in S$ be one of these. Then $\{1\} \subset S$ is a proper subset. Accordingly the product subset
is not open. For if it were, it would have to be the union of product subsets that contain the total set $S$ in at least one entry, which by construction it is not.
Write $Disc(\{0,1\})$ for the the discrete topological space with two points. Write $\underset{n \in \mathbb{N}}{\prod} Disc(\{0,2\})$ for the product topological space of a countable set of copies of this discrete space with itself (i.e. the corresponding Cartesian product of sets $\underset{n \in \mathbb{N}}{\prod} \{0,1\}$ equipped with the Tychonoff topology induced from the discrete topology of $\{0,1\}$).
Then consider the function
which sends an element in the product space, hence a sequence of binary digits, to the value of the power series as shown on the right.
One checks that this is a continuous function (from the product topology to the Euclidean metric topology on the closed interval). Moreover with its image $\kappa\left( \underset{n \in \mathbb{N}}{\prod} \{0,1\}\right) \subset [0,1]$ equipped with its subspace topology, then this is a homeomorphism onto its image:
This image is the Cantor space as a subspace of the closed interval.
(projections are open maps)
For $\{X_i\}_{i \in I}$ a set of topological spaces, the for each $j \in I$ the projection out of their product space into the $j$th component
is an open map.
Since images preserve unions (this prop.) it is sufficient to check that the image under $p_j$ of a base open? is still open. But base opens in the product topology by definition are, in particular, products of open subsets.
The product topological space construction from def. is the limit in Top over the discrete diagram consisting of the factor spaces, hence the category theoretic product.
For proof see at Top – Universal constructions.
The Tychonoff theorem states that the product space of any set of compact topological spaces (with its Tychonoff topology) is itself compact.
(one-point compactification intertwines Cartesian product with smash product)
On the subcategory $Top_{LCHaus}$ of Top on the locally compact Hausdorff spaces with proper maps between them, the functor of one-point compactification (Prop. )
sends Cartesian products (product topological spaces) to smash products of pointed topological spaces, hence constitutes a strong monoidal functor, in that there is a natural homeomorphism:
This is briefly mentioned in Bredon 93, p. 199. The argument is spelled out in: MO:a/1645794/, Cutler 20, Prop. 1.6.
The singular homology of product topological spaces is informed by
examples of universal constructions of topological spaces:
$\phantom{AAAA}$limits | $\phantom{AAAA}$colimits |
---|---|
$\,$ point space$\,$ | $\,$ empty space $\,$ |
$\,$ product topological space $\,$ | $\,$ disjoint union topological space $\,$ |
$\,$ topological subspace $\,$ | $\,$ quotient topological space $\,$ |
$\,$ fiber space $\,$ | $\,$ space attachment $\,$ |
$\,$ mapping cocylinder, mapping cocone $\,$ | $\,$ mapping cylinder, mapping cone, mapping telescope $\,$ |
$\,$ cell complex, CW-complex $\,$ |
The Tychonoff topology is named after A. N. Tychonoff.
Terence Tao, Notes on products of topological spaces (pdf)
Florian Herzig, Product topology (pdf)
John Terilla, Notes on categories, the subspace topology and the product topology 2014 (pdf)
Last revised on October 24, 2021 at 08:32:32. See the history of this page for a list of all contributions to it.