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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
4
votes
Hurewicz theorem on mappings that lower dimension
In Engelking's Theory of Dimensions, Finite and Infinite, Thm 3.3.10 (p. 200) proves the more general result
If $f: X \to Y$ is a closed mapping from a normal space $X$ to a weakly paracompact normal …
3
votes
Accepted
The Tychonov cube $X^X$ of a compact space $X$ is a compact semigroup with the composition o...
To check that $f \to f \circ g$ is continuous in $f$ as a map $X^X \to X^X$ for a fixed $g \in X^X$: take a net $f_i \to f$ ($i \in I$, some directed set) in $X^X$ converging to $f \in X^X$. This mea …
2
votes
Compact spaces whose compactness does not come from a product of compact spaces
For me the most basic compact spaces are $\{x_n: n \in \Bbb N\} \cup \{x\}$ when $x_n \to x$, (the countable cofinite space is a special case), of course all finite spaces, and all ordered topological …
2
votes
Accepted
Metrizability of topology of compact convergence
According to Engelking (exercise 3.4E, which is based on a paper by Arens):
If $C(X,\Bbb R)$ (with the compact-open topology and $X$ Tychonoff) is first countable, then $X$ is hemicompact.
A Hau …
8
votes
Accepted
Which compact metrizable spaces have continuous choice functions for non-empty closed sets?
It's an old (1981) theorem by Jan van Mill and Evert Wattel (see this paper) that a compact space has a continuous selection iff it is orderable. (So has a linear order whose order topology is the top …
5
votes
Accepted
A classification of $G_{\delta\sigma}$ zero-dimensional spaces?
This paper by Van Mill from 1981 gives a characterisation of $\Bbb Q \times \Bbb P$ (where $\Bbb P$ is a common notation for the irrationals) in Thm 5.3:
If $X$ is separable metrisable and zero-di …
3
votes
Is the lexicographic ordering on the unit square perfectly normal?
For a compact Hausdorff $X$ it is equivalent that $X$ is hereditarily Lindelöf or that $X$ is perfectly normal. Sketch: $X$ is hereditarily Lindelöf implies that every open set is an $F_\sigma$ (as $X …
0
votes
Accepted
A closed subset $B$ of the Hilbert cube such that $\operatorname{Int}(B) = \emptyset$ and $B...
$A=\{0\} \times [-1,1]^{\Bbb N}$ works as an example, e.g.
For more info on $Z$-sets in $Q$, see Infinite-dimensional Topology by Jan van Mill.
5
votes
Accepted
Is a cofinite topology for a set with cardinality between $\aleph_{0}$ and $2^{\aleph_{0}}$ ...
A continuous non-constant function from $[0,1]$ into $X$ with the cofinite topology exists iff $[0,1]$ has a partition into $\le |X|$ many disjoint closed non-empty subsets.
This question discusses …
1
vote
Accepted
Lower semicontinuity of a multi-valued map $F:X\to 2^Y$ in term of net
You're probably right about the statement of the theorem.
Indeed, many textbooks (especially older ones) call nets "generali(s/z)ed sequences"
I think it should say:
Given a net $x: \mathcal{I} \ …
5
votes
Accepted
Continuity of the Restriction Map Between Function Spaces
If $K$ is compact in $Z$ and $U$ open in $Y$, then $K$ is still compact as
a subset of $X$ as well (compactness is absolute, or maybe use that $i[K]$ is compact where $i: Z \to X$ is the continuous c …
2
votes
Piecewise-metrizability problems from Willard's Topology
Your corrected version of exercise 2:
Suppose $X$ is $T_4$ and $X=\bigcup \mathcal{U}$ where all $U \in \mathcal{U}$ are open and metrisable, and $\mathcal{U}$ is locally finite. Then $X$ is metr …
4
votes
Accepted
Are compactifications of completely $T_{4}$ spaces completely $T_{4}$?
The obvious (to me) counterexamples are $\beta \mathbb{N}$ and $\beta \mathbb{R}$ ( the Čech-Stone compactifications) which are non-completely normal compactifications (classic fact, see Engelking or …
2
votes
Two notions of boundedness in metrizable topological vector space
For boundedness of sets the statement is false. The Wikipedia quote is for linear operators.
A counterexample for sets: $X=\mathbb{R}^\omega$ in the product topology is a metric locally convex TVS. …
3
votes
Accepted
Is there a second countable topological space, which can not be equipped with a finite borel...
A simple solution: if $X$ is second countable, let $D=\{d_n : n =1,2,3,\ldots\}$ be a dense subset of $X$ and define $$\mu(A)= \sum_{n:d_n \in A}\frac{1}{2^n}$$ for all subsets of $X$.
Then clearly …