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
analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
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For metric spaces with their metric topology it is true that sequentially compact metric spaces are equivalently compact metric spaces. The analogous statement fails for more general topological spaces: for them, being sequentially compact in general neither implies nor is implied by being compact (see the counter-examples here).
But the failure of this equivalence is not due to a deficit in the concept of convergence but in the concept of sequences and sub-sequences. If the latter are generalized to nets and sub-nets (beware that the definition of sub-nets is slightly non-obvious), then the analogue of the statement remains true in generality: A topological space is compact precisely if every net in it has a sub-net that convergence.
(compact spaces are equivalently those for which every net has a converging subnet)
Assuming excluded middle and the axiom of choice, then:
A topological space $(X,\tau)$ is compact precisely if every net in $X$ has a sub-net that converges.
We break this up into lemmas 1 and 2:
(in a compact space, every net has a convergent subnet)
Let $(X,\tau)$ be a compact topological space. Then every net in $X$ has a convergent subnet.
Let $\nu \colon A \to X$ be a net. We need to show that there is a subnet which converges.
For $a \in A$ consider the topological closures $Cl(S_a)$ of the sets $S_a$ of elements of the net beyond some fixed index:
Observe that the set $\{S_a \subset X\}_{a \in A}$ and hence also the set $\{Cl(S_a) \subset X\}_{a \in A}$ has the finite intersection property, by the fact that $A$ is a directed set. Therefore this prop. implies from the assumption of $X$ being compact that the intersection of all the $Cl(S_a)$ is non-empty, hence that there is an element
In particular every neighbourhood $U_x$ of $x$ intersects each of the $Cl(S_a)$, and hence also each of the $S_a$. By definition of the $S_a$, this means that for every $a \in A$ there exists $b \geq a$ such that $\nu_b \in U_x$, hence that $x$ is a cluster point of the net.
We will not produce a sub-net
that converges to this cluster point. To this end, we first need to build the domain directed set $B$. Take it to be the sub-directed set of the Cartesian product directed set of $A$ with the directed neighbourhood set $Nbhd_X(x)$ of $x$
on those pairs such that the element of the net indexed by the first component is is contained in the second component:
It is clear $B$ is a preordered set. We need to check that it is indeed directed, in that every pair of elements $(a_1, U_1)$, $(a_2, U_2)$ has a common upper bound $(a_{bd}, U_{bd})$. Now since $A$ itself is directed, there is an upper bound $a_3 \geq a_1, a_2$, and since $x$ is a cluster point of the net there is moreover an $a_{bd} \geq a_3 \geq a_1, a_3$ such that $\nu_{a_{bd}} \in U_1 \cap U_2$. Hence with $U_{bd} \coloneqq U_1 \cap U_2$ we have obtained the required pair.
Next take the function $f$ to be given by
This is clearly order preserving, and it is cofinal since it is even a surjection. Hence we have defined a subnet $\nu \circ f$.
It now remains to see that $\nu \circ f$ converges to $x$, hence that for every open neighbourhood $U_x$ of $x$ we may find $(a,U)$ such that for all $(b,V)$ with $a \leq b$ and $U \supset V$ then $\nu(f(b,V)) = \nu(b) \in U_x$. Now by the nature of $x$ there exists some $a$ with $\nu_a \in U_x$, and hence if we take $U \coloneqq U_x$ then nature of $B$ implies that with $(b, V) \geq (a,U_x)$ then $b \in V \subset U_x$.
Assuming excluded middle, then:
Let $(X,\tau)$ be a topological space. If every net in $X$ has a subnet that converges, then $(X,\tau)$ is a compact topological space.
By excluded middle we may equivalently prove the contrapositive: If $(X,\tau)$ is not compact, then not every net in $X$ has a convergent subnet.
Hence assume that $(X,\tau)$ is not compact. We need to produce a net without a convergent subnet.
Again by excluded middle, then by this prop. $(X,\tau)$ not being compact means equivalently that there exists a set $\{C_i \subset X\}_{i \in I}$ of closed subsets satisfying the finite intersection property, but such that their intersection is empty: $\underset{i \in I}{\cap} C_i = \emptyset$.
Consider then $P_{fin}(I)$, the set of finite subsets of $I$. By the assumption that $\{C_i \subset X\}_{i \in I}$ satisfies the finite intersection property, we may choose for each $J \in P_{fin}(I)$ an element
Now $P_{fin}(X)$ regarded as a preordered set under inclusion of subsets is clearly a directed set, with an upper bound of two finite subsets given by their union. Therefore we have defined a net
We will show that this net has no converging subnet.
Assume on the contrary that there were a subnet
which converges to some $x \in X$.
By the assumption that $\underset{i \in I}{\cap} C_i = \emptyset$, there would exist an $i_x \in I$ such that $x \neq C_{i_x}$, and because $C_i$ is a closed subset, there would exist even an open neighbourhood $U_x$ of $x$ such that $U_x \cap C_{i_x} = \emptyset$. This would imply that $x_J \neq U_x$ for all $J \supset \{i_x\}$.
Now since the function $f$ defining the subset is cofinal, there would exist $b_1 \in B$ such that $\{i_x\} \subset f(b_1)$. Moreover, by the assumption that the subnet converges, there would also be $b_2 \in B$ such that $\nu_{b_2 \leq \bullet} \in U_x$. Since $B$ is directed, there would then be an upper bound $b \geq b_1, b_2$ of these two elements. This hence satisfies both $\nu_{f(e)} \in U_x$ as well as $\{i_x\} \subset f(b_1) \subset f(b)$. But the latter of these two means that $\nu_{f(b)}$ is not in $U_x$, which is a contradiction to the former. Thus we have a proof by contradiction.