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I have reread the proof and it's completely correct. The idea is that $\omega_1 \times \beta\mathbb{N}$ maps perfectly onto a non-normal space, and normality is preserved under perfect maps. Tamano's …

This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is here. It's in fact not that hard, the fact for a product of copies of 2 point discrete spaces already implies …

Some results from Engelking's dimension theory book: If $f: X \mapsto Y$ is a closed, continuous and surjective function between normal spaces $X$ and $Y$, and $\forall y \in Y: | f^{-1}[{y}] | \le k …

You already found a (classical) counterexample: the double arrow ($[0,1] \times \{0,1\}$, ordered lexicographically), which is even compact and separable. There is however a nice metrization theorem f …

The property "Every uncountable set has a limit point" is related to the Lindelöf property (every open cover has a countable subcover). For metrisable spaces these notions are equivalent, and in gener …

So we start with $(X, \mathcal{T})$, a $T_1$ space in which every point is a $G_{\delta}$, as witnessed by open sets $U_n(x)$, $n \in \mathbb{N}$, $x \in X$. W.l.o.g. we can take these sets to be decr …

Lemma: if $K$ and $L$ are (order) convex in a linearly ordered set $X$, and $x$ is in $K \cap L$, then $K \cup L$ is convex as well. Proof: suppose $a < b$ are in $K \cup L$ and $c$ lies in $(a,b)$. …

A space is Lindelöf iff it satisfies condition A and is metaLindelöf (every open cover has a point-countable refinement), so one could say the difference is metaLindelöfness.

No to all your questions. There are lots of countable non-metrizable spaces: an easy one is $\mathbb{N}$ in the cofinite topology, which can be written as a countable (disjoint) union of singletons (w …

For an infinite Hausdorff space the diversity of a space is the number of homeomorphism types of non-empty open sets, so if all non-empty open sets are homeomorphic, the space is said to be of diversi …

Suppose $X$ is metaLindelöf, and $A \subseteq X$ is hereditarily Lindelöf and dense. Let $\mathcal{U}$ be an open cover of $X$ and let $\mathcal{V}$ be a point-countable refinement of $\mathcal{U}$. …

The Hilbert cube $[0,1]^\omega$. Just split the coordinates in two disjoint infinite sets. Most standard sequence spaces like $\ell^\infty, \ell^2$ will similarly work.

Indeed, check the paper by Gauld. Your (4) implies his condition hemicompact. His example at p15, that you saw refutes the just separable condition (7). Note that (1)-(6) imply imply metrisability for …

In Engelking's Theory of Dimensions, Finite and Infinite, Thm 3.3.10 (p. 200) proves the more general result If $f: X \to Y$ is a closed mapping from a normal space $X$ to a weakly paracompact normal …

Let $(y_n)$ be a sequence in $Y$. Let $A$ be the subset of $Y \times \mathbf{R}$ of all points $(y_n, \frac{1}{n})$ for $n \in \mathbf{N}$, and let $B$ be its closure. Then $\pi_2[B]$ is closed in $\m …