Joe

@AeroMain27: There's not actually an easy answer to that question – in formalised mathematics (set theory, type theory, etc.), the answer will depend on the details on your foundational system. I could try to write more, but I'm not sure how helpful it would be. Have you studied foundations of mathematics before?

To be honest, I always found it kind of funny that there are authors who attempt to write "general" algebra books. The idea that someone who go page by page through a book like Dummit and Foote, learning group and ring theory, then representation theory, then algebraic geometry,... it just strikes me as a little odd

Anyone who has tried learning algebra from Lang has my deepest sympathies

@anankElpis: Hi, thanks for commenting on my question about field extensions. Did you have any luck finding a reference for the claim that $|\operatorname{Hom}_K(L,\Omega)|=2^{[L:K]_s}$? I'd be quite interested in seeing it, since most algebra books that I am familiar with focus on finite extensions

Suppose $K$ is field, $i:K\to L$ is a finite algebraic extension, and $j:K\to\Omega$ is an algebraic closure of $K$. It is known that the set of $K$-morphisms $L\to\Omega$ is at most $[L:K]$. Is this result still true if $L$ is an infinitely generated algebraic extension (in which case $[L:K]$ is an infinite cardinal)?

To be honest I don't like the definition of an algebra as being a certain subspace of $C(X,\mathbb R)$. It's like defining a vector space to $\mathbb F^n$ for some field $\mathbb F$ and $n\in\mathbb N$. However, if you are going to use an abstract definition of an algebra $A$ (such as the one suggested by Xander), you should probably also assume that the bilinear product $A\times A\to A$ is associative. Otherwise, notation like $c_1\dots c_n$ doesn't even make sense

@psie: Can you start with your definition of "algebra", since there are about 17 different definitions in the literature?

That's a funny result

Every time a mathematician applies Zorn's Lemma they are secretly appealing to transfinite recursion.

If $A$ is an integral domain, we say that $A$ is a normal domain if $A$ is integrally closed in $\operatorname{Frac}(A)$. Do normal domains have anything to do with normal extensions of fields? I presume that they do not, but I just wanted to check...

@XanderHenderson What did the empty set ever do to you? ;)

@psie Yes

@psie Yes

I suppose, since the quarternions have a normed structure, and a division ring structure, you could just define $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \, .$$But you would have to say whether $\frac{f(x+h)-f(x)}{h}$ means $h^{-1}[f(x+h)-f(x)]$ or $[f(x+h)-f(x)]h^{-1}$.

@Jakobian I'm curious: what definition of "differentiable" are you using so that the only differentiable functions over the quarternions are constants?

@anankElpis In fact, arguably it is easier to give examples of continuous non-differentiable functions in the complex case compared to the real case. For example, complex conjugation is continuous but not differentiable anywhere. In the real case, showing that there exists a continuous but nowhere differentiable function is much harder, and all of the ways I have seen that do it involve the use of nontrivial theorems, such as the Baire Category Theorem.

Thanks for letting me know. I'll try checking your proof again tomorrow

You are just jealous of how all of our algebras are unital, associative, and commutative ;)

You should move over to algebraic geometry. None of the spaces are Hausdorff, and you might even go insane

This is taken from Locally Compact Groups by Markus Stroppel

(Here, $\mathcal C(Y,Z)$ is the set of continuous maps $Y\to Z$ with the compact-open topology.)

@psie: By the way, I have not studied Arzela-Ascoli for a while, and I am aware that there are several versions of this theorem, but does this theorem answer whether you need Hausdorffness?

It doesn't help that there are about four different definitions of what an "algebra over a field" is

Other fun facts: $\mathbb Q$ is the initial object in the category of Archimedean ordered fields, whereas $\mathbb R$ is the final object.

Fun fact (for a perverse definition of "fun"): the Dedekind-completion of any Archimedean ordered field is isomorphic to $\mathbb R$.

I waded my way through the construction of $\mathbb R$ using Dedekind cuts about 4 years ago, when I was first learning real analysis. Never again ;)

If you have defined $\mathbb Z$ as an explicit subset of $\mathbb R$, then it has the advantage of making $\mathbb Q$ easy to define: it is $\{a/b\mid a,b\in\mathbb R\text{ and }b\neq0\}$. (Or, I guess you could define $\mathbb Q$ as the smallest subfield of $\mathbb R$, but that doesn't feel very analysis-y.)

Then, you can define $\mathbb Z$ as the set $\{a-b \mid a,b\in\mathbb N\}$, for example.

@Seeker When proving statements such as this (which are completely "obvious" from a naive perspective), it's very important for you to say exactly which results you are allowed to use. So, how are you defining $\mathbb Q$ and $\mathbb Z$? (For example, if you have defined $\mathbb R$ as a complete ordered field, then $\mathbb N$ can be defined as the smallest (with respect to inclusion) subset $A\subseteq \mathbb R$ such that $0\in A$ and $n+1\in A$ whenever $n\in A$.)

Sorry, edited my last comment

For example, if $R$ is the ring of one-variable polynomial functions with coefficients in $\mathbb Q$, then the natural map $\mathbb Q[x]\to R$ is injective (it has a trivial kernel; that is, if a polynomial vanishes everywhere, then it is the zero polynomial)

When working over an infinite field, each polynomial function is induced by one and only one polynomial

Well, if you are working over an infinite field like $\mathbb Q$, then there is no significant difference between polynomials and polynomial functions

@leslietownes Oh, that is much simpler than what I was trying to do. Very nice. Everything here is characteristic zero, since a field extension of $\mathbb Q$ has characteristic zero

I meant to write $g(i\pi,e)=f(-\pi,e)f(\pi,e)=0$.

Wait, I think my argument had a few typos in it...

To answer my own question: suppose that $e$ and $\pi$ were algebraically dependent, say $f\in\mathbb Q[x,y]$ is a nonzero polynomial such that $f(\pi,e)=0$. Then, let $g\in\mathbb Q[x,y]$ be given by $g(x,y)=f(ix,y)f(-ix,y)$ (though it is a bit annoying to check that the coefficients lie in $\mathbb Q$). We have $g(\pi,e)=f(i\pi,e)f(-i\pi,e)=0$, hence $e$ and $i\pi$ are algebraically dependent; contradiction.

I mean, if we know that $e$ and $i\pi$ are algebraically independent over $\mathbb Q$, does it obviously follow that $e$ and $\pi$ are algebraically independent over $\mathbb Q$?

I have a (probably trivial) question: Schanuel's Conjecture implies that $\mathbb Q(e,i\pi )$ has transcendence degree $2$ over $\mathbb Q$. Apparently, this in turn implies that $e$ and $\pi$ are algebraically independent, but I'm not sure how to show this.

@Jakobian: There are still trivial counter-examples, I think. Start with an infinite family $\mathcal V$ of pairwise disjoint sets. Pick a nonempy set $A\in\mathcal V$, and pick a proper subset $A'$ of $A$. Take $\mathcal U=\mathcal V\cup\{A'\}$.

Now, I understand that there is room for debate about this, but this is why I think some people would argue that stating the Baire Category Theorem as a result about metric spaces is somewhat misleading. There are other theorems in analysis where we do actually care about what the metric is (i.e. we are not treating the metric up to equivalence of norms). For example, any theorem that uses the word "isometry" is fundamentally about metric spaces, not about metrisable spaces

@XanderHenderson To answer your question directly: because it arguably makes it clearer that the result is fundamentally topological in nature, rather than about metrics per se. To apply the Baire Category Theorem to a space $X$, you only need to know that there exists a metric on $X$ which gives it its topology; what the metric actually is is irrelevant.

Indeed, let $X$ be a nonempy completely metrisable space. By definition, there exists a metric $d$ on $X$ which induces the topology on $X$, and gives $X$ the structure of a complete metric space. Hence $X$ is not a countable union of nowhere dense sets. Now all you need to do is remark that the notion of "nowhere dense" depends solely on the topology on $X$, not on its metric.

@XanderHenderson I now realise your earlier comment was facetious but I'm not sure I agree with your later one. Once you know that the Baire Category Theorem is true for complete metric spaces, it trivially follows that is true for completely metrisable spaces.

Perhaps some set-theoretic topologists and analysts, too

As for who studies metrisable spaces: descriptive set theorists and (some) general topologists.

@XanderHenderson I'm quite sympathetic to the idea of stating results in terms of metrisable spaces rather than metric spaces (probably not a good idea if you are teaching undergraduates). The advantage is that makes it clear when a theorem about metric spaces is purely topological in nature, or when we only care about a metric space up to equivalence of norms. The disadvantage is that it is a little cumbersome if every proof has to start with "Let $d$ be a metric on $X$ which induces its topology..."

But someone with a more algebraic background might tell them that they are working in the wrong category, and the correct statement is "A nonempty completely metrisable space cannot be written as a countable union of nowhere dense sets".

As pointed out by user21820 in this answer, the use of the word "let" in ordinary mathematics corresponds to two very different meanings in mathematical logic. This is rather unfortunate from a pedagogical point of view.