@niloderoock Your definition of "extremely degrading" doesn't agree with mine.

@CarloBeenakker Here's a naïve failure of $\delta$. Let $H$ be the Heaviside step function, with $H' = \delta$, and note that, under pointwise multiplication of functions, $H^n = H$ for any $n \in \mathbb{N}$. Then $\delta = H' = (H^2)' = 2HH' = 2H\delta$, but also $\delta = (H^3)' = 3H^2 H' = 3 H \delta$. So $H\delta = 0$ and therefore $\delta = 0$. As a distribution, $H^n$ doesn't exist, and certainly doesn't equal $H$.

You can get something like a "sphere with centre everywhere and circumference nowhere" in the framework of non-standard analysis by taking a sphere with radius the reciprocal of an infinitesimal. If you move any finite amount, you have only moved an infinitesimal proportion of the radius away from the centre.

@AliTaghavi Actually, that was needlessly complicated. Let $X$ be an uncountable set, considered as a discrete space, and $Y$ it's one-point compactification. Then $Y$ is not first countable (at the new point because $X$ is not countable), but is compact, Hausdorff and every sequence has a convergent subsequence.

@AliTaghavi If Martin's axiom holds, then every product of $< 2^{\aleph_0}$ sequentially compact spaces is sequentially compact. So if the continuum hypothesis is false but Martin's axiom holds, then $2^{\aleph_1}$, with the product topology, is compact, Hausdorff, every sequence has a convergent subsequence, but is not first countable. There may well be a ZFC example, but I don't know one off the top of my head.

Kripke and Platek developed their theory not as an alternative foundation motivated by some supposed deficiency in ZF, but as a tool for proving theorems in higher recursion theory. See Kripke's abstracts.

@WillBrian "Meatspace" for the real world, in opposition to "cyberspace" was popularized by William Gibson in his science fiction novels, e.g. Neuromancer. He disclaims inventing the term. It then became a commonly-used hacker retronym, like "snail mail" for ordinary mail (in contradistinction to e-mail).

This holds for probability measures because the set of probability measures is equicontinuous. Beware that on the entire space of bounded signed measures these are three distinct weak topologies.

As Wojowu suggests, use $\subseteq$. The intersection of a family of subdcpos is a subdcpo, so the set of subdcpos is a complete lattice, and therefore a dcpo.

@JamesHanson For any large cardinal axiom A, one cannot prove con(ZF) => con(ZF + A) unless ZF is inconsistent, so that doesn't say much.

@LeeMosher YMMV = "your mileage may vary" and is an ancient Usenet-ism. (Or was that the joke?)

@LSpice Users need 50 reputation to comment on other people's posts, including questions, so Joey W didn't have the ability to comment. See here.

@SamSanders In Turing's correction (starting on the middle of the second page) he does not attribute the pointing out of the error to Brouwer, although he mentions the law of the excluded middle, and, in a footnote, Brouwer's work on the reals. The only person to whom he gives credit for pointing out errors is Paul Bernays.

I know an answer to the question in the title at least. The category of compact Hausdorff spaces $\mathrm{CHaus}$ is monadic over $\mathrm{Set}$ by the usual forgetful functor and its left adjoint the Stone-Čech compactification. Gelfand duality makes $\mathrm{CHaus}$ dual to the category of commutative unital C$^*$-algebras $\mathrm{CC}^*$. Then the unit ball functor $\mathrm{CC}^* \rightarrow \mathrm{Set}$ is also monadic, in fact $\aleph_1$-accessibly so - the free commutative C$^*$-algebra on $X$ is $C(\mathbb{D}^X)$ where $\mathbb{D}$ is the closed complex unit disc.

The bounded weak-* topology is easier to characterize than the weak-* topology, therefore (if you have a locally convex topology on a normed space in which the unit ball is compact, and the topology is the finest linear topology agreeing with the topology on the ball, this space is a dual Banach space with the bounded weak-* topology). Of course, the topology on the unit ball is a piece of structure, not something intrinsic to the normed space structure, because of the existence of Banach spaces with non-isometric preduals.

@NikWeaver The statement of your answer omits the necessary assumption in the question and Jochen's comment that $\tau$ be coarser than $\sigma(X',X'')$, which has led the OP of this question astray. If $X$ is infinite-dimensional, then the bounded weak-* topology $\tau$ on $X'$ is locally convex, strictly finer than $\sigma(X',X)$, and agrees on the unit ball. Of course $\tau$ is not coarser than $\sigma(X',X'')$ (it's finer if $X$ is reflexive, and incomparable otherwise).

Nik Weaver's answer has an extra assumption that is stated in the question and Jochen Wengenroth's comment (to Nik's answer) - that the topology on $X'$ is also coarser than $\sigma(X',X'')$ (i.e. the weak (not weak-) topology of $X'$). Without this assumption, there is the bounded weak- topology on $X'$ which is strictly finer than $\sigma(X',X)$ in the infinite-dimensional case. In fact it is the finest unique finest linear topology agreeing with $\sigma(X',X)$ on the unit ball.

If $X$ is a Banach space equipped with a locally convex topology $\mathcal{T}$ in which $B_X$ is compact, then $X$ is the dual space of the linear maps $X \rightarrow k$ that are continuous when restricted to $B_X$ (where $k$ is the base field $\mathbb{R}$ or \mathbb{C}$). The "locally convex" is necessary by a counterexample of J. W. Roberts.