All Questions
14 questions
2
votes
0
answers
406
views
Complete topological groups in which all subgroups are closed
My previous question has been answered by YCor; so I am asking a new one with a reasonable additional assumption. See the previous question for the background and motivation.
General question: does ...
2
votes
1
answer
49
views
Is any submetrizable linear topology linearly submetrizable?
Let $E$ be a vector space. A topology $\tau$ on $E$ is called (linearly) submetrizable if there is a (linear) metrizable topology $\pi$ on $E$ which is weaker than $\tau$, i.e. $\pi\subset\tau$.
Is ...
2
votes
1
answer
213
views
Parametrization of topological algebraic objects
There are several results of the following form: if an algebraic objects is endowed with a topology (or rather uniformity) which is somehow compatible with the algebraic structure, this uniformity is ...
1
vote
0
answers
48
views
Neighborhoods of idempotents in topological inverse semigroups
In a topological group, for any neighborhood $U$ of the origin, there is another such neighborhood with the property that $V.V\subseteq U.$ I conjecture a similar property for topological inverse ...
1
vote
0
answers
81
views
Morphism in commutative square strict?
Let $G,H$ be topological groups and $f:G\rightarrow H$ a continuous group homomorphism.
Then $f$ is said to be strict if $G/\mathrm{Ker}(f) \cong \mathrm{Im}(f)$ is an isomorphism of topological ...
8
votes
1
answer
829
views
Topological groups in which all subgroups are closed
General question: does there exist a nondiscrete topological group $G$ such that all subgroups of $G$ are closed? Or, does there exist a nondiscrete topological vector space $V$ such that all vector ...
4
votes
2
answers
263
views
Sufficent condition for strict morphism of normed vector spaces
Let $K$ be a non-archimedean field of char 0 and a morphism $f:V \rightarrow W$ of normed $K$-vector spaces given. The map $f$ is said to be strict if $V/\ker(f)$ with the quotient topology is ...
3
votes
1
answer
107
views
Is each cometrizable space a subspace of a cometrizable topological group?
Following Gruenhage we call a topological space $X$ cometrizable if $X$ admits a weaker metrizable topology such that every point $x\in X$ has a (not necessarily open) neighborhood base consisting of ...
2
votes
1
answer
352
views
The completeness of spaces of continuous functions with the compact-open topology
For a Tychonoff space $X$ let $C_k(X)$ denote the space of continuous real-valued functions on $X$, endowed with the compact-open topology.
Problem. Is the space $C_k(X)$ Polish if it is Polishable ...
3
votes
1
answer
201
views
Is each closed subgroups of $\mathbb R^\omega$ isomorphic to a Tychonoff product of locally compact Abelian groups?
It is known that any closed linear subspace of $\mathbb R^\omega$ is topologically isomorphic to $\mathbb R^n$ for some $n\in\omega$.
Problem 1. Is each closed subgroup of $\mathbb Z^\omega$ (or ...
4
votes
0
answers
156
views
Basic calculus on topological fields
Let $K$ be a a topological field (I am mainly interested in the cases when K is either an ordered field or a valued field, e.g. $K = \mathbb Q$ or $ \mathbb Q_p$).
1) Let $f: K^n \to K$ be a ...
6
votes
1
answer
249
views
Extending the topology on a set to the group/vector space it generates
The multiplicative group $\Bbb Q^+$ can be viewed as a $\Bbb Z$-module. To see this, note that any rational can be decomposed into the form
$2^{n_2} \cdot 3^{n_3} \cdot 5^{n_5} \cdot ...$
The tuple ...
3
votes
1
answer
727
views
Tensor product of topological abelian groups with the reals
Given an abelian group A, the tensor product $A \otimes R$, where R are the reals, is naturally an R-vector space.
Now suppose that A is a topological abelian group (if necessary, we can assume it to ...
6
votes
3
answers
2k
views
Sequential topological vector spaces
Since I'm dealing with the distinction between sequential continuous and continuous maps at the moment I came to ask myself once again what can be said about spaces where these two notions agree (...