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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
4
votes
0
answers
152
views
Two other variants of Arhangel'skii's Problem
This question is a follow up to another question of mine, which turned out to be easy (for background on Arhangel'skii's Problem see Arhangel'skii's problem revisited). Recall that a space is linearly …
10
votes
0
answers
239
views
Arhangel'skii's problem revisited
One of the most well-known problems in set-theoretic topology is Arhangel'skii's question of whether there exists a Lindelöf Hausdorff space with "points $G_\delta$" (meaning, every point is the inter …
3
votes
0
answers
121
views
A space with independent tightness
Recall that the tightness of a topological space $X$ is defined as the least cardinal $\kappa$ such that for every non-closed subset $A$ of $X$ and every point $x \in \overline{A} \setminus A$, there …
9
votes
2
answers
437
views
Convergence properties in dense subsets of $\omega^*$
The space $\omega^*$, the remainder of the Cech-Stone compactification of the integers, fails to have all convergence-type properties known to me.
Sequentiality. (As a matter of fact $\omega^*$ does …
6
votes
0
answers
162
views
Free sequences and the cardinality of a topological space
One way of formulating Arhangel'skii's celebrated theorem about the cardinality of Lindelof first-countable spaces is the following (due to Arhangel'skii and Shapirovskii). For every Hausdorff space $ …
19
votes
1
answer
652
views
A large separable space of countable tightness
Is there a ZFC example of a Tychonoff space $X$ such that:
$X$ is separable.
$X$ has countable tightness (that is, a subset of $X$ is closed if and only if it contains the closure of each one of its …
8
votes
1
answer
182
views
Are all monotonically normal manifolds of dimension at least two metrizable?
Alan Dow and Frank Tall recently proved the consistency of the statement Every hereditarily normal manifold of dimension at least two is metrizable.
See: Dow, Alan; Tall, Franklin D., Hereditarily nor …
6
votes
0
answers
168
views
On the cellularity of the $G_\delta$-topology
Given a topological space $X$, let $X_\delta$ be the topology on $X$ generated by the $G_\delta$ subsets of $X$. Let $c(X)$ be the cellularity of $X$, that is, the supremum of cardinalities of familie …
10
votes
1
answer
414
views
A variant of the Moore-Mrowka problem
A space $X$ is said to be sequential if whenever $A \subset X$ is not closed then $A$ contains a sequence converging to a point outside of $A$.
A space $X$ is said to have countable tightness if for …
5
votes
0
answers
112
views
Is there a homogeneous compactum where non-empty $G_\delta$s have non-empty interior?
A space $X$ is called an almost $P$-space if $Int(G) \neq \emptyset$ for every non-empty $G_\delta$ subset $G \subset X$.
Every $P$-space (that is, a space where $G_\delta$s are open) is an almost $P …
6
votes
0
answers
155
views
Is there a Lindelof $P$-space which is not discretely generated?
A space $X$ is:
Lindelof if every open cover for $X$ has a countable subcover.
A $P$-space if every $G_\delta$ subset of $X$ is open.
Discretely generated if for every non-closed set $A \subset X$ an …
4
votes
0
answers
78
views
Is there an $L$-space whose square is selectively $d$-separable?
An $L$-space is a hereditarily Lindelof regular space which is not separable.
A space is $d$-separable if it contains a dense set which is the countable union of discrete sets.
An $L$-space can't …
7
votes
1
answer
254
views
What's the minimal weight of a maximal space?
A non-empty topological space without isolated points is called maximal if every finer topology on that space has at least an isolated point. The existence of a (Hausdorff) maximal space is a simple c …
3
votes
1
answer
321
views
Is there a linearly Lindelof space which is not weakly Lindelof?
Recall that a space is:
"Lindelof", if every open cover has a countable subcover.
"Linearly Lindelof", if every open cover which is linearly ordered by $\subseteq$ has a countable subcover.
"weakly …
6
votes
0
answers
151
views
Countably compact non-compact perfect spaces
Recall that a space is countably compact if every infinite set has an accumulation point. A space is perfect if every closed set is a countable intersection of open sets. One of the classical applicat …