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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- …

How about this example: $[0,1]^A$ with the product topology, where $A$ is uncountable. Then every nonempty $G_\delta$ set is uncountable. Since your sets $f^{-1}(x)$ are $G_\delta$ sets, this has yo …

Ulm invariants. Surely someone still knows this? Given $f \colon A \to A$ and $g \colon B \to B$, is there a bijection $\phi \colon B \to A$ such that $f(\phi(x))=\phi(g(x))$? There is a system of …

There is also this result... $\displaystyle\sum_{a \in A} \frac{1}{a}$ diverges if and only if the span of $\{x^a | a \in A\}$ is dense in the continuous functions on an interval. (I guess you have …

Your criterion is (half of) uniform convergence. As commented, not every function can be uniformly approximated by Riemann integrable functions. I say "half" because you wrote $f(x) -R(x) < \epsilon$ …

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 …

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 …

We will use the Kuratowski–Ryll-Nardzweski selection theorem: Let $(\Omega, \mathscr{F})$ be a measurable space. Let $E$ be a Polish space. Let $\Gamma$ be a set-valued function from $\Omega$ to $E$ …

The video series "Dimensions"... http://www.dimensions-math.org/ [Available for free download, viewing on-line, or purchase on DVD.] Episodes 7 and 8 on fibrations contain computer graphics intende …

I we omit "path" in the formulation, then it cannot be done. I guess that is why it is formulated this way. Suppose the plane could be written as the union of two totally disconnected sets. Inter …

Every metric space of nonmeasurable cardinality is realcompact. [1] 15.24. Thus, if there are no measurable cardinals, then every metric space is realcompact As you noted...To get a Banach space t …

Field $\mathbb Q$ with the usual absolute value $|\cdot|$ from the real numbers. Two norms on $\mathbb Q^2$ ... $$ \|(x,y)\|_1 = |x|+|y| $$ and $$ \|(x,y)\|_2 = \left|\,x+\sqrt{2}\;y\,\right| $$ In b …

No, not in general. My metric space is the disjoint union of uncountably many copies of $\mathbb R$. $$X = \bigsqcup_{t \in T} X_t$$ where $T$ is uncountable and $X_t = \mathbb R$ for all $t$. The …

(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 …

first answer. As stated ($X, Y$ merely metric spaces), NO. (Remark: we may as well take $D=Y$ and $f$ the identity on $D$: if we can do that case, then we can apply it to get the general case.) Let …