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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.

33 votes
4 answers
2k views

Connectedness in the language of path-connectedness

Is there a topological space $(C,\tau_C)$ and two points $c_0\neq c_1\in C$ such that the following holds? A space $(X,\tau)$ is connected if and only if for all $x,y\in X$ there is a continu …
Dominic van der Zypen's user avatar
31 votes
2 answers
1k views

Is $\mathbb{R}\cong\text{Cont}(X,Y)$ for some non-trivial spaces $X,Y$?

For topological spaces $X,Y$ let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y.$ We endow $\text{Cont}(X,Y)$ with the topology inherited from the product topology on $Y^X.$ A …
Dominic van der Zypen's user avatar
25 votes
3 answers
1k views

What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?

This is a follow-up on the following question. Let $\text{End}(X)$ denote the endomorphism monoid of a topological space $X$ (that is, the collection of all continuous maps $f:X\to X$ with composition …
Dominic van der Zypen's user avatar
20 votes
2 answers
540 views

$\kappa$-homogeneous topological spaces

Let $\kappa>0$ be a cardinal and let $(X,\tau)$ be a topological space. We say that $X$ is $\kappa$-homogeneous if $|X| \geq \kappa$, and whenever $A,B\subseteq X$ are subsets with $|A|=|B|=\kappa$ …
Dominic van der Zypen's user avatar
19 votes
3 answers
2k views

Non-homeomorphic spaces such that taking away a point makes them homeomorphic

Are there topological spaces $X,Y$, each having more than $2$ points, satisfying the following two properties? $X\not\cong Y$, and there is a bijection $\varphi: X\to Y$ such that for all $x\in X$ t …
Dominic van der Zypen's user avatar
19 votes
3 answers
1k views

"Anti" fixed point property

Let $(X,\tau)$ be a topological space. If $f:X\to X$ is continuous, we say $x\in X$ is a fixed point if $f(x) = x$. The space $(X,\tau)$ is said to have the anti fixed point property (AFPP) if the on …
Dominic van der Zypen's user avatar
14 votes
3 answers
409 views

$T_2$ topologies that are "as disjoint as possible"

Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$? (That is, the in …
Dominic van der Zypen's user avatar
14 votes
2 answers
1k views

Hausdorff spaces with trivial automorphism group

Is the singleton space the only Hausdorff space $X$ such that the set of automorphisms $\varphi: X\to X$ equals $\{\textrm{id}_X\}$?
Dominic van der Zypen's user avatar
13 votes
1 answer
673 views

Strongly rigid Hausdorff spaces

A space $(X,\tau)$ is called rigid if $\textrm{Aut}(X)=\{\textrm{id}_X\}$. We say $(X,\tau)$ is strongly rigid if for every continuous map $f:X\to X$ we have that $f = \textrm{id}_X$ or $f$ is constan …
Dominic van der Zypen's user avatar
13 votes
1 answer
631 views

$T_2$-spaces where all non-empty open sets are homeomorphic

We say that a $T_2$-space $(X,\tau)$ has homeomorphic open sets if every non-empty open set $U\subseteq X$ endowed with the subspace topology is homeomorphic to $(X,\tau)$. The rationals with the Euc …
Dominic van der Zypen's user avatar
12 votes
1 answer
642 views

Universal maps between topological spaces

Let $X,Y$ be topological spaces. We call a continuous map $u:X\to Y$ universal if for every continous map $f:X\to Y$ there is $x\in X$ such that $f(x) = u(x)$. If $u:X\to Y$ and $v:Y\to Z$ are univer …
Dominic van der Zypen's user avatar
12 votes
2 answers
1k views

Are countable dense subspaces of $\mathbb{R}^n$ homeomorphic to ${\mathbb Q}^n$?

Let $n\geq 1$ be an integer and suppose $S\subseteq {\mathbb R}^n$ is countable and dense. Do we have $S \cong {\mathbb Q}^n$ where both sets carry the topology inherited from the Euclidean topology o …
Dominic van der Zypen's user avatar
12 votes
3 answers
2k views

Minimal Hausdorff

A Hausdorff space $(X,\tau)$ is said to be minimal Hausdorff if for each topology $\tau' \subseteq \tau$ with $\tau' \neq \tau$ the space $(X,\tau')$ is not Hausdorff. Every compact Hausdorff space i …
Dominic van der Zypen's user avatar
11 votes
Accepted

Topological conditions forcing continuity

It's possible to find a sufficient condition without putting any restraint on the spaces $X$ or $Y$. Let $Z$ be the Sierpinski space, that is $Z$ has $\{0,1\}$ as a base set, and $\tau = \{\emptyset, …
Dominic van der Zypen's user avatar
10 votes
1 answer
475 views

Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$?

Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$, where $\mathbb{R}^\omega$ is given the product topology, and $x\in\mathbb{R}^\omega $?
Dominic van der Zypen's user avatar

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