I'm stuck on the following real analysis problem. Show that a set $E$ of real numbers has Lebesgue measure zero if and only if there is a countable collection $\{I_{k}\}_{k=1}^{\infty}$ of open intervals for which each point in $E$ belongs to infinitely many of the $I_{k}$'s and $\sum_{k=1}^...

Let ${\mathcal D}(f)$ be the set of points at which $f$ is not continuous. The correct result is: Theorem: Let $f:[a,b] \rightarrow {\mathbb R}$ be bounded. Then the following are equivalent: (a) $\;f$ is Riemann integrable on $[a,b].$ (b) $\;{\mathcal D}(f)$ is a countable union of closed Jordan...

Is a set of Lebesgue measure zero necessarily a countable union of sets of Jordan content zero? This was a question posed by a student in my undergraduate analysis course. I asked an analyst colleague about this and he did not have an answer off the top of his head. Here are a few thoughts about...

Yes. The proof I know is a little roundabout. Let $M$ be your manifold. It is locally compact and Hausdorff, so it has a one-point compactification $M^*$ which is compact Hausdorff. Now $M^*$ is again second countable (see One point compactification is second contable), and (locally) compac...

Recall that a link is an embedding of some number of circles in space up to isotopy. (Think "a knot but maybe multiple pieces.") Three oriented links form a skein triple if they have identical diagrams except for at one place, where they differ as in this image: (This image should be familiar to...

$(X, \tau_X) $ and $(Y, \tau_Y) $ be two topological spaces. $\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $. Question : Does this implies $(Y, \tau_Y) $ is compact? Notation: $Y^X$: Set of all functions from $X$ to $Y$. $C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$...