Alessandro Codenotti

Left and right are difficult in all fields. Some people call the the left uniformity what other people call the right uniformity

Reason #999 to be doing set theory rather than category theory

They are easy in set theory

@Thorgott I have no clue what exact functors do, but those are easy

@BalarkaSen no worries! Still doing some kind of geometry? Are you doing a PhD now? I'm doing well, I'm also still doing maths, I'm a postdoc in Bologna now

@Jakobian I never realized that I've always seen the $K_\sigma$ notation exclusively in descriptive set theory books/papers and not really in topology ones. Funny

Long time no see @BalarkaSen, how's life? What are you up to these days?

@Anacardium How do you define convex in a space which is not a TVS?

@Thorgott Chain complexes and cochain mplexes

A quick google search suggests he's working in Sweden now

I don't know whether he's still in Bonn. He had a temporary position at the time

Oberdieck

But I do distinctly remember that during algebraic geometry I the professor felt comfortable assuming that everyone knows what a (co)representable functor is, but felt it necessary to remind people what "Hausdorff" means!

Makes sense, I wouldn't know, I only took Topologie I (also with Lück) when I was there as far as (algebraic) topology courses go

Of course even the most basic topology course in Bonn is heavily algebraic topology oriented :P

Any (real) vector space will have infinitely many elements, regardless of its dimension

The dimension of the vector space will be the same as the number of elements in the set

Not that there will be finitely many functions

No, it only says that the set of functions will be a finite dimensional vector space

There is no issue, functions on this space are exactly the same as real numbers, hence a one dimensional vector space

I'm taking arbitrary functions $\{x\}\to\Bbb R$

But the real numbers are a one dimensional vector space, so in particular a finite dimensional one, as claimed

Or in other words there is a 1-1 correspondence between real-valued functions on the space with one point and real numbers

So the function is entirely determined by its value on the only available point

Sure, but the space you're starting from has only point

@imbAF What kind of functions are there on a set with one element?

I'm thinking about weird subsets of the reals though

I stick mostly to continua nowadays, haven't really looked at weird spaces in a while

@Jakobian I had to leave earlier, but I don't really know what $F$ or $F'$ spaces are

A cozero set is just the complement of the zero set of a continuous function right?

Sure

Yes, why?

And this is a contradiction because by construction $g$ agrees with $f$ on $A$, and $f$ is known by hypothesis to be continuous on $A$+

@psie a function $g$ is continuous on $A$ if and only if for every sequence $(x_n)\subseteq A$ with limit $x_0$ also in $A$, $g(x_n)\to g(x_0)$. Here you built a sequence as in the hypothesis, but which does not converge to $g(x_0)$, so $g$ is not continuous on $A$

For example a constant function

Its own upper bound in which ordered set?

@Thorgott no.

I would have probably used "glueing $F$-spaces along a $P$-set" but that's because I never use the adjunction space name

It's clear what you mean in my opinion

But of course all commonly studied theories have a countable (finite more often than not) language

A stupid example is that if you have a language with uncountably many constant symbols $(c_i)_{i\in I}$ and axioms $c_i\neq c_j$ for every pair $i\neq j$ in $I$, then all models of this theory will have cardinality at least $|I|$

Lowenheim-Skolem guarantees the existence of a theory in every cardinality not smaller than that of the language

@RyderRude exactly

@AlessandroCodenotti Of course if you run this algorithm on a non theorem, it will never find it in the list and just run forever

For any first order theory in a countable language, but there are also theories in uncountable languages

And then the algorithm to recognize whether a formula is a theorem is to run the algorithm producing all theorems as above and stop as soon as you find the formula in the list

In that case you can write down al algorithm that lists all theorem in order (fix some computable enumeration of the axioms, go though it and apply all possible inference rules)

Semidecidability is a very weak notion though, every first order theory with recursively enumerable axioms is semidecidable

It always halts on theorems, but might run forever on non theorems

Well you can run the algorithm again with the negation of your statement to check