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I definitely expect that there is much more than one good answer. But, here is one that one can get easily by just patching together several classical facts: The category of compact Hausdorf topolog …

I confirme my comment : $X$ is the stone-cech compactification of a discrete space if and only if $X$ is compact, haussdorf, extremally disconected, and has a dense set of open points. here is a ske …

So the short answer is that there is no non-empty projective locales for essentially any reasonable class of epimorphisms you can think of (except maybe proper maps). The problem is that there exists …

If the absence of adjoints is what worries you, you can consider this to be a two-step process - and I would argue that in practice this is the case in the vast majority of cases: One first generalize …

Let me expand a bit my comment as this is a rather subtle property. As I said any locally compact Hausdorff topological space is a strongly hausdroff locally compact locales. (and under the axiom of …

I have thought about this recently, and here is I think the best constructively valid statement one can extract from Brouwer fixe point theorem (framework : internal logic of an elementary topos, real …

I like to call this result the localic Baire category theorem, and it plays essentially the same role as Baire category theorem: it lets you "construct" object by showing that some spaces are non-empt …

When $X$ is compact, it is discrete (hence not compact unless it is finite). For any function $f:X \to \{0,1\}$, then both $K_0 = f^{-1}(0)$ and $K_1=f^{-1}(1)$ are compact subset of $X$, and so the s …

Don't pick a convergent subsequence: pick a point in $ \cap_{n \in \mathbb{N}} \overline{\{ x_n,x_{n+1}, \dots \}}$ (the overline meaning closure) which is non empty as an intersection of decreasing f …

The answer is No. Rougly, because it is not a good idea to look at continuous $\mathbb{C}$ valued function on a space which is not completely Haussdorff as completely haussdorf is exactly the hypothes …

I claim that the following result have constructive* proof: 1) Let $f : [0,1] \rightarrow \mathbb{R}$ be a uniformly continuous function such that $f(0)\leqslant 0$ and $f(1) \geqslant 0$ then (as a …

For short, the exponential $(X,Y)$, characterized by the usual universal properties: morphisms from any locale $Z$ to $(X,Y)$ are functions from $X \times Z$ to $Y$, exists for all $Y$ if and only if …

Let me first clarify some confusion in the comments to the original question. To be clear : I'm not at all saying the persons making them were confused, as far as I can tell all the comments were corr …

Note : In order to define the alexandrov compactification of $X$ you have to take the algebra of function converging at infinity (not just those converging to $0$) If by a compactification of $X$ you …

I'm assuming that by "compact" you only mean which satisfies the finite cover properties, and not compact and Hausdorff. in this case the following produces a counterexample: Let $X_1$ be $\mathbb{R} …