Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
8
votes
Accepted
"Immovable" topological spaces
No. Let $X$ be the disjoint union of the real line and one isolated point. The quotient by "movability" collapses the real line to one point, so the quotient is a discrete space of two points, which …
9
votes
Accepted
Adjoining a new isolated point without changing the space
In my answer to https://mathoverflow.net/questions/26414 I descibed a somewhat simpler-looking example than Nik's, but proving that it works may be harder. Take two copies of $\beta\mathbb N$ and glu …
3
votes
Accepted
Union of proximally connected sets
Let $\{X,Y\}$ be a partition of the union of the family $S$, and let p be a point in the intersection of $S$ (which you've assumed is nonempty). Without loss of generality, p is in $X$. But, since a …
9
votes
The difference between a sequential space and a space with countable tightness
All three notions, "countably tight," "sequential," and "Frechet-Urysohn," say that each point $p$ in the closure of a set $A$ can be "approached in some countable way" by points from $A$. The differ …
7
votes
An ultrafilter and a partition
Non-principal ultrafilters with the property in the question for all partitions are called "selective" or "Ramsey" ultrafilters. The "Ramsey" terminology comes from the following connection, due to K …
4
votes
Accepted
closed meagre sets
For Q3, the answer is yes. Think of the Cantor set as consisting of the points that have a base-3 expansion containing only 0's and 2's. Now take the subset of those points where the 2's occur only …
4
votes
closed subset of weakly lindelof
No. Consider the space whose points are all sequences of 0's and 1's of length $\leq\omega$. Visualize it as the binary tree plus "limits" for all paths through the tree, and topologize it accordingl …
8
votes
Accepted
Uniqueness and existence of topology for a given convergence class of nets
It is not possible to have two topologies on $X$ with the same nets converging to the same points. To prove it, consider any two distinct topologies $T$ and $T'$ on $X$, and suppose, without loss of g …
40
votes
When factors may be cancelled in homeomorphic products?
In the category of topological spaces, you can't in general even cancel the two-point discrete space. The only counterexample I know is a bit complicated though: Start with the disjoint union of two …
8
votes
extracting a convergence subnet from a sequence which is Cauchy on every bounded subset of $...
Let the complete bounded metric space be the unit ball of the Hilbert space $l_2$, and let $\{e_i:i\in\mathbb N\}$ be an orthonormal basis. Consider the following sequence $(x_n)$, which goes from ea …
7
votes
Limits of rearranged sequences along ultrafilters
As Joel David Hamkins commented, this is likely to be a cardinal characteristic. Let me define (I hope temporarily) the relevant characteristic $\mathfrak{dfpmk}$ (named after the OP and the author o …
6
votes
A question on minimal idempotent ultrafilter on N^2
Since $(\mathbb N-\{0\})^2$ is a 2-sided ideal in $\mathbb N^2$, it follows that $\beta((\mathbb N-\{0\})^2)$ is a 2-sided ideal in $\beta(\mathbb N^2)$ and therefore contains all of the latter semigr …
8
votes
Periodic point-free maps and free ultrafilters.
Although the question has already been thoroughly answered, it might be worthwhile to point out that the result can be separated into the combinatorial "meat", which doesn't involve ultrafilters, and …
3
votes
Nonmetrizable uniformities with metrizable topologies
Let $X$ be an infinite set. Take the uniformity $\mathcal U$ generated by those equivalence relations on $X$ that have only finitely many equivalence classes. The induced topology on $X$ is discrete …
3
votes
Injective Function on a Dense Set
Joel has completely answered the question, but let me add another example, with a bigger failure of injectivity of $g$. Let $D$ be the set of points in the plane of the form $(\frac1n,\sin n)$ for po …