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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
6
votes
0
answers
169
views
Topological spaces for which $w(X)\leq |X|$ holds
Let $w(X) = \inf\{|\mathcal{B}| : \mathcal{B} \text{ is a base for }X\}$ be the weight of topological space $X$.
For metric spaces and locally compact spaces we have inequality $w(X)\leq |X|$.
This in …
3
votes
2
answers
129
views
Countable zero-sets are $C$-embedded?
I was browsing Gillman and Jerison for known relations between zero-sets, $C$-embedded sets and so on.
The spaces I'm considering are $T_{3.5}$.
There are two properties that pseudocompact spaces have …
3
votes
0
answers
227
views
Adjunction of pointed maps is a homeomorphism?
What interests me the most is if the case of exponential law is true under the assumptions claimed for example on nlab: if $X, Y$ are Hausdorff and $Y$ is locally compact, then $F^0(X, F^0(Y, Z))\cong …
5
votes
0
answers
269
views
$T_1$ paratopological group having a dense commutative subgroup is commutative
I'm learning about topological groups from Arhangelskii and Tkachenko "Topological groups and related structures" and this is one of the exercises there.
A paratopological group is a group with topolo …
22
votes
1
answer
663
views
Are $\beta \mathbb{Q}$ and $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ homeomorphic?
The canonical inclusion $\beta\mathbb{Q}\setminus \mathbb{Q} \hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of $\beta\mathbb{Q}\setminus \mathbb{Q}$. Even so, this doesn't nec …
3
votes
0
answers
79
views
Pseudocompactness, countable compactness and locally finite open covers
Let $(P_1)$ be the property: Every locally finite open cover of $X$ has finite subcover.
Let $(P_2)$ be the property: Every locally finite open cover of $X$ is finite.
Let $(P_3)$ be the property: Eve …
3
votes
1
answer
113
views
Perfectly normal but not collectionwise normal space in ZFC
In the article A Perfectly Normal, Locally Compact, Noncollectionwise Normal Space Form $\lozenge^\ast$ by Daniels and Gruenhage (I presume "form" is a typo and it should be "from", but it's like that …
3
votes
1
answer
71
views
Even covers and collectionwise normal spaces
We call $X$ strongly collectionwise normal if the set $\mathcal{U}_\Delta$ of all neighbourhoods of the diagonal $\Delta_X$ of $X\times X$ is a uniformity. This is equivalent to the property that for …
0
votes
0
answers
20
views
Equivalent conditions for $z$-embeddability
I am looking for where this specific theorem of Blair is originally located:
Theorem. Let $S\subseteq X$, the following are equivalent:
$S$ is $z$-embedded
If $A, B\subseteq S$ are disjoint zero-set …
4
votes
1
answer
233
views
Mysior plane is not realcompact
Let $X = \mathbb{R}^2$ with $(x, y)\in X$ for $y\neq 0$ isolated and $(x, 0)$ having neighbourhood basis of the form $$U_n(x) = \{(x, y) : y\in (-1/n, 1/n)\}\cup \{(x+y+1, y) : 0 < y < 1/n\}\cup \{(x+ …
2
votes
1
answer
82
views
LCH spaces $X$ such that if $Y$ is a perfect image of $X$, then $Y$ is zero-dimensional
I am looking for locally compact Hausdorff spaces $X$ with the following property:
If $f:X\to Y$ is a perfect map onto locally compact Hausdorff space $Y$, then $Y$ is zero-dimensional.
One can see …
1
vote
0
answers
89
views
Well-embedded type property for bounded functions
According to @Tyrone the term well-embedded set was first used in Measures on Metacompact Spaces by W. Moran.
In the article Extensions of Zero-sets and of Real-valued Functions by R. Blair and A. Hag …
3
votes
1
answer
172
views
Extensions of bounded uniformly continuous functions
Let $X$ be a uniform space, $S\subseteq X$ and $f:S\to \mathbb{R}$ bounded uniformly continuous, then there exists a uniformly continuous extension of $f$ to $X$. (Katětov, 1951)
I am looking for pr …
1
vote
1
answer
86
views
Real exponentiation in the quotients of rings of continuous functions by prime ideals
Consider the ring $C = C(X) = C(X; \mathbb{R})$ of continuous functions $f:X\to \mathbb{R}$ where $X$ is a Tychonoff space. This is naturally a lattice ordered ring by setting $f\geq 0$ iff $f(x)\geq …
6
votes
2
answers
286
views
Reference: If $X$ is metrizable, then $X$ is realcompact iff $|X|$ is non-measurable
Note: What I call a measurable cardinal seems to be non-standard among set theorists, and should be called a $\sigma$-measurable cardinal.
I know that a discrete space is realcompact iff its non-measu …