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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 ...
erz's user avatar
  • 5,529
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 ...
erz's user avatar
  • 5,529
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 ...
Bumblebee's user avatar
  • 1,093
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 ...
KKD's user avatar
  • 473
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 ...
Leonid Positselski's user avatar
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 ...
Leonid Positselski's user avatar
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 ...
KKD's user avatar
  • 473
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 ...
Taras Banakh's user avatar
  • 41.9k
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 ...
Taras Banakh's user avatar
  • 41.9k
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 ...
Taras Banakh's user avatar
  • 41.9k
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 ...
Antongiulio Fornasiero's user avatar
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 ...
Mike Battaglia's user avatar
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 ...
AlexE's user avatar
  • 2,998
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 (...
Johannes Hahn's user avatar