In intuitionistic mathematics, a non-constant function from $\mathbb R$ to $\{ 0 , 1 \}$. Many classical theorems can be proved to fail intuitionistically by showing that they imply this or ...

I would call this an internal category in the category of topological spaces and continuous maps.

There are several interpretations of the original question, but one is, why focus on open sets rather than closed sets? I have an unusual answer. Suppose you want to do constructive mathematics. (Don'...

Let $C$ and $D$ both be the category of finite-dimensional (say real) vector spaces and invertible linear maps between them, let $F$ be the identity, and let $G$ take a vector space to its dual. ($G$ ...

Although it's a little wordy, there is a method of formalizing things that avoids these theorems. To be sure, theorems such as $\{ \{ \} , \{ \{ \} \} \} \in \{ \{ \} , \{ \{ \} , \{ \{ \} \} \} \}$...

In the theory of electromagnetism, the classical Stokes Theorem moves between the differential and integral forms of two of Maxwell's four equations; see https://en.wikipedia.org/wiki/Stokes%...

I can't give you an answer that fully addresses what Lawvere and Rosebrugh were thinking, since I haven't asked them. (If you want to ask them, Rosebrugh runs a category-theory mailing list at https:/...

Mike Shulman has stack semantics, an application of stacks to logic. This is basically sheaf semantics, a now standard application to logic of sheaves (far from their own origin in geometry), except ...

Localic completion is awesome and is the way to answer your question as asked (which requests a locale). However, a simpler answer for getting the correct (internal) set of $p$-adic numbers (the one ...

A dimensionful real-valued quantity takes values in a $1$-dimensional real vector space (a ‘line’) rather than in the space of real numbers as such. Given two such quantities taking values in the ...

It looks like the term ‘ideal-supporting algebra’ was written by me and survived slightly more than a decade on Wikipedia without being altered. (Well, somebody added a hyphen, a change that I agree ...

I don't think that there is anything interesting about this constructively, as long as we limit ourselves to finite sets in the strictest sense. The proof that a winning strategy exists gives us no ...

There's this nLab page: http://www.ncatlab.org/nlab/show/math+blogs and some other lists that it links to.

Characterization is too strong; there are many models of $\operatorname{ETCS}$, so its axioms don't characterize $\operatorname{Set}$ either. But I'll interpret your question as asking how the axioms ...

An adjunction between posets is precisely a Galois connection. (Paul has characterized these in the case where the posets are both complete.) English Wikipedia; nLab.

I want to write down a proof that comes naturally, in a way. This proof assumes that you can form the quotient of a set modulo an equivalence relation, but does not require powersets. (So it works ...

Here is another, completely different answer. As Carlo indicated, one can use units in which Planck's constant is $1$. This is no arbitrary choice, but one dictated by fundamental physics. ...

As has already been remarked by Lasse Rempe-Gillen, you need to know what a Cauchy net is, say from a uniform structure. But since you want $\kappa$ to be discrete [edit: I had ‘complete’ here once] ...

Like Terry Tao, I find the transience of slides to be a problem. This is one reason why I stopped using slides as such and began using a single continuous-scroll page for each topic. I lecture from ...

I'm interested in the differentials-based approach advocated by Dray and Manogue at Bridging the Vector Calculus Gap. This is for multivariable calculus, but they do discuss the one-variable version (...

Well, I've decided to go ahead and use ‘with enough points’. There are a lot of reasons to restrict to $T_0$ spaces, over and above reasons to restrict to sober spaces, and at least within that ...

I'm afraid that I don't like your proposed proof. You derive a bound on $f(x)$, namely $\epsilon + f(y)$, but this is not fixed. Although you may choose any positive $\epsilon$ you wish (which then ...