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Partial answer: if the space is zero-dimensional then any two disjoint open sets have disjoint closures, so the space is extremally disconnected. And then closures of open sets are clopen, hence your …

How about Bing's example: The cartesian product of a certain nonmanifold and a line is $E^4$. The `other' factor is the dog-bone decomposition of three-space.

Consider $X\cap A$ and $Y\cap A$, starting from a partition $\lbrace X,Y\rbrace$ of $A\cup B$. If both intersections are nonempty we are done, as $(X\cap A)\delta(Y\cap A)$. Otherwise, $A\subseteq X$, …

An answer is: completely regular plus countable pseudocharacter, the latter means that points are $G_\delta$-sets. In completely regular spaces a point is a $G_\delta$-set iff it is the zero-set of a …

Assume the complement of $S$ in $\mathbb{R}^n$ is not connected, say $A$ and $B$ are relatively closed and disjoint in $\mathbb{R}^n\setminus S$ (and nonempty of course); let $O$ be the complement of …

Another answer was given by Aarts and De Groot in Complete regularity as a separation axiom Canadian Journal of Mathematics 21 1969 96–105. Frink's condition sets things up for a proof a la Urysohn's …

All it takes is a countable, not first-countable, Tychonoff space, say a countable dense subset, $D$, of the Cantor cube $2^{\mathfrak{c}}$. For every point $(d,e)$ off the diagonal there is a continu …

The two sets are essentially the same: the map that sends every $f\in I^{\beta X}$ to its restriction is a bijection; the two topologies are, in general, not the same. The compact-open topology on $I^ …

Let $\mathcal{F}$ be this partition. As noted above it must be uncountable. We may as well assume it lives on the interval $(0,1)$ and add the set $ \lbrace 0, 1\rbrace$ to each member to obtain a fam …

For regular spaces the implication is false in general: let $A$ be regular and such that all continuous real-valued functions on it are constant (see Problem 2.7.17 in Engelking's General Topology). T …

The Niemytzki plane is weakly Lindelöf (the open upper half plane is a dense Lindelöf subspace); the $x$-axis is an uncountable closed and discrete subspace.

Yes, take, for example the Sorgenfrey plane $P$. A standard example of a non-normal space. It is separable and its anti-diagonal $\lbrace (x,-x):x\in\mathbb{R}\rbrace$ is closed and discrete, so Jones …

Consider the $F$-space $\omega^* $ (the Cech-Stone remainder of $\omega$). Under the Continuum Hypothesis there is no point whose complement is $C^* $-embedded. On the other hand it is also consistent …

Here's a counterexample to 1. Let $T$ be the Tychonoff plank, i.e., the product $(\omega_1+1)\times(\omega+1)$ with the point $\langle\omega_1,\omega\rangle$ removed. Consider the set $\omega\times …

If you want a closed discrete set take the metric hedgehog with $\kappa$ many spines, where $\kappa$ is the desired cardinality. If $\kappa<\mathfrak{c}$ replace $[0,1]$ by a countable connected Hausd …