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This is false. Example. Let $Y$ be a totally disconnected topological space in which no point is open; for example $Y = \mathbb Z_p$. Construct $X$ from $Y$ by adding a closed point in the closure of …

Let my write up my comment as an answer. The easiest argument I know uses sheaf cohomology. It requires working through a bit of theory, but ultimately gives a very flexible tool for proving all sorts …

No such space exists. We actually get the stronger statement that every isomorphism $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(\mathbf R)$ is induced by an isomorphism $X \stackrel\sim …

This is certainly not true in general. Of course in the scheme-theoretic setting where $f$ is locally of finite presentation, this is Chevalley's theorem [Tag 054K], but this is far from a purely topo …

This is not true, already in $\mathbf R^2$. Indeed, if $V$ is finite-dimensional, then the Euclidean topology is the coarsest topology for which all linear¹ maps $V \to \mathbf R$ are continuous: choo …

These are the only ones. Write $x \preceq y$ if and only if $x = x \wedge y$ if and only if $y = x \vee y$, and write $\leq$ for the usual partial order on $\mathbf R$. To avoid some case separations, …

As noted in this answer to a previous question of yours, for any subvariety $W \subseteq \mathbb A_\mathbb R^n$, we have $$\dim_{\mathbb R} W(\mathbb R) \leq \dim W$$ (where $\dim W$ denotes the dimen …

This is false, already for spectra of valuation rings. Recall that ideals in a valuation ring $R$ correspond to ideals in the value group $\Gamma$, which can be any totally ordered abelian group. More …

Proposition. Let $X$ be a topological space. Then $X$ is a compact object if and only if $X$ is a finite discrete space. Before giving the proof, we state an easy lemma. Lemma. Suppose $ …

Here is an example: Example. Let $g$ be a continuous strictly increasing function such that $\lim_{x \to -\infty} g(x) = -1$ and $\lim_{x \to \infty} g(x) = 1$; for example $$g(x) = \tfrac{2}{\pi}\ar …

As Wojowu noted in the comments, one should really look at $T_1$ topological spaces. Consider the functors \begin{align*} G\!: \mathbf{Top}_{T_1} &\leftrightarrows \mathbf{Cond}_\kappa:\!F\\ X &\mapst …

Note that Esquisse d'un programme is a speculative text, and moreover we cannot ask the author for clarification, so the best you can get is a speculative answer. That said, I do think that Grothendie …