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An example, I hope. Write $X \sqcup Y $ for disjoint union, say a set made up of two disjoint closed parts, one homeomorphic to $X$ and one homeomorphic to $Y$. And write $\bigsqcup_{i} X_i$ for a d …

Q: Why do we use certain operations rather than others? For example the usual addition for the real numbers? A: Because, if we used some other addition it would not be what we call "the real num …

One result of this kind: if $X$ is zero-dimensional and $f$ is at most $n$ to one, then $\dim f(X) \le n-1$. This is if and only if... Given $Y$ of dimension $m$, there is a zero-dimensional $X$ and …

$C$ is simply a connected complete separable metric space with more than one point. Now the union of countably many closed sets of topological dimension zero must again have dimension zero. But I sup …

This is the main step of the proof that the plane (in this case, the square) has topological dimension 2. You can find a proof (as elementary as I could make it) in my text Measure, Topology, and Fra …

As noted, goes back to de Rham. Found also in: Georges de Rham, "Sur quelques courbes définies par des equations functionnelles". Univ. e Politec. Torino. Rend. Sem. Mat. 16 1956/1957 101–113. …

There are points in the Cantor set that are not endpoints of any of the removed intervals. For example $1/4$ is such a point.

There are, in fact, uncountably many gaps between the intervals of your construction. Indeed, the complement $F$ of your open set is (homeomorphic to) a Cantor set. You are correct that it contains …

What does "hole" mean? All the empty spaces are connected to each other, so probably you should say there is one hole. LINK

If $U= (-\infty,0) \times \mathbb R \times \mathbb R \times \dots$, then $\mathrm{cl}(U) = (-\infty,0] \times \mathbb R \times \mathbb R \times \dots$, and the complement of this is $(0,+\infty) \tim …

Perhaps look for "uniform spaces" ... this is a generalization of metric spaces, with no need to use real numbers in the definition. Wikipedia has a page on uniform spaces. Many textbooks on point-s …

comment I found HERE A space is called screened if every open covering has a $\sigma$-disjoint open refinement Do you think this is equivalent to weakly Lindelof: every open covering has a $\si …

What does this example do ... All spaces are on set $\{1,2,\dots\}$. Space $X_n$ has topology that makes $\{1,2,\dots,n\}$ discrete and $\{n+1,\dots\}$ indiscrete. Of course $X_n$ is compact non- …

(too long for a comment to Pete's answer) Garrett Birkhoff was my Ph.D. advisor. Let me provide a few remarks of a historical nature. From a 25-year-old Garrett Birkhoff we have: Abstract 355, "A n …

Bockstein's theorem Bockstein, M., Un théorème de séparabilité pour les produits topologiques, Fundam. Math. 35, 242-246 (1948). ZBL0032.19103. This is the case of a product $\prod_{t \in T} X_t$ wher …