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There is a famous question of Arhangel'skii: Is there a non-discrete extremally disconnected topological group? The general problem is still open, but the separable case was solved a few years ago b …

This relation was introduced (I don't know if for the first time) in the 1984 paper Bijectively related spaces I: Manifolds by P. H. Doyle and J. G. Hocking. As the title indicates, two spaces that a …

There is a book called Function Spaces with Uniform, Fine and Graph Topologies by Robert A. McCoy, Subiman Kundu, Varun Jindal. I haven´t read it but it has a chapter called Cardinal Functions and Cou …

If $X$ is hereditarily Lindelof then any open subset of $Y$ is Lindelof and therefore it is the union of countably many basic (for a given predetermined base for $Y$) open sets. Hence The $\sigma$-alg …

Any compact homogeneous hereditarily separable space has size at most $\mathfrak{c}$ (this is a result of Ismail). Thus any compact homogeneous separable space of bigger size provides a counterexample …

It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ …

If $B$ is a complete boolean algebra (i.e. if $S(B)$ is extremally disconnected) then Property $P$ is equivalent to "$(X,\subseteq)$ is a well-order". On the other hand if $B$ is countable (and I supp …

Since any function from $X$ into $[0,1]$ is bounded, I suppose you want to characterize the continuous functions from $X$ into $[0,1]$. If you choose any sequence $a=\langle a_n : n \in \omega \rangl …

A map $f:X \to Y$ is ring-like if for every point $x \in X$ and every pair $U$ and $V$ of open neighborhoods of $x$ and $f(x)$ respectively, there is an open $W$ such that $f(x) \in W \subseteq V$ an …

There exists a metrizable topological group $H$ such that $H \setminus \{e\}$ is rigid (see Theorem 6.1 in van Mill´s paper: A topological group having no homeomorphisms other than translations). Ex …

The answer is yes for countable graphs: Fix an infinite graph $G$ and a bijective homomorphism $f:G \to G$. Define $c:[G]^2 \to 2$ as $c(\alpha,\beta)=1$ if $\{f\alpha, f\beta\} \in E(G)$ and $c(\alp …

No. Just looking at countably many generators we can produce a continuum of pairwise disjoint clopen subsets of $X$. Moreover, since $|A|=2^{\aleph_0}$, we have that $2^{\aleph_0} \leq c(X) \leq d(X) …

Note that replacing "well-ordered" by "linearly-ordered" produces an equivalent property since any linear order contains a cofinal well order. Such spaces were called lob-spaces and studied by S.W. Da …

The answer is no. A space is called resolvable if it contains two disjoint dense subspaces. Clearly $X$ is resolvable if and only if $\chi(X)=2$. Lets prove by induction on $n \geq 2$ that if $\chi(X …

The answer is no. In the paper "A separable normal topological group need not be Lindelöf" (General topology and its applications, 1976), Hajnal and Juhász use the continuum hypothesis to construct a …