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If one wants constant functions to always be continuous, then one must necessarily have the empty set and the whole space to be open. From a category theory perspective, it is the continuous function …

One could phrase local compactness using neighbourhood bases as well (in the Hausdorff case, at least) if desired: once one has one precompact open neighbourhood, one automatically has a whole neighb …

Let $X$ be an uncountable discrete space with a distinguished element $0$. We view the product space $X^X$ as the space of maps $\phi: X \to X$. The set $$ E := \{ \phi \in X^X: \phi(\phi(0)) = 0 \} …

By Fubini's theorem, the sum of the reciprocals of the primes is equal to $\int_1^\infty \frac{\pi(x)}{x^2}\ dx$, where $\pi(x)$ is the number of primes less than x. The prime number theorem tells us …

I believe the answer is "no". The key lemma is: Lemma. Let $f: [c,d] \to {\bf R}$ be smooth, let $I$ be a compact subinterval of $(c,d)$, $q$ be an interior point of $I$, let $n \geq 1$, and let $\ …

Let $E$ be a set of the claimed form. Call a direction $\omega \in S^1$ a limit direction of $E$ if there exists a sequence $p_n$ of points in $E$ going to infinity whose argument goes to $\omega$, o …

Yes. It suffices to show that if one has a sequence $\vec v^{(n)} = (v^{(n)}_1,\dots,v^{(n)}_m) \in E^m$ whose Gram matrix $(\langle v^{(n)}_i, v^{(n)}_j \rangle)_{i,j=1,\dots,m}$ converges to a Gram …

One can think of a topology on a space $X$ as abstracting all the "stable" information (or "physical measurements") one can say about a state $x$ in $X$ (i.e. the open neighbourhoods of $x$ in $X$). …

The topological entropy I am assigning to a sequence $f$ (Definition 2) is not directly related to the similar-sounding concept of topological sequence entropy (Definition 1), but is instead related t …

The textbook presentation of a topology as a collection of open sets is primarily an artefact of the preference for minimalism in the standard foundations of the basic structures of mathematics. Thi …

One example I know of is the Kazdan-Warner identity: if $(M,g)$ is a Riemannian surface, with scalar curvature $R$, and $X$ is a conformal Killing vector field, then $\int_M R \operatorname{div}(X)\ d …

I think the following construction gives a counterexample. It stems from the observation that the Playfair axiom is quite weak in the case where all lines only have three points (it produces some pai …

The surface area $|\partial S|$ of a (bounded, smooth) body $S$ is the derivative of the volume $|S_r|$ of the $r$-neighbourhoods $S_r$ of $S$ at $r=0$: $$ |\partial S| = \frac{d}{dr} |S_r| |_{r=0}.$ …