Are there any infinite descending chains of closed subgroups of the general linear group over a field? More specifically, is my argument ok? Can you fill in some of the details?
Prop: No.
Proof: If we define $|H| = ( \dim(H), [ H:H_0 ] )$ where $H_0$ is the connected component of the identity and if we order these “cardinalities” lexicographically, then I claim that in any chain $H < K$ that $|H| < |K|$. Since the ordered pairs of non-negative integers has no infinite descending chains, the same is then true for the closed subgroups. Proof of the claim: Consider $H_0, K_0$. I claim $H_0 \leq K_0$ [ but I'm not sure why ]. Since these are both connected, either $H_0 = K_0$ or $\dim(H) =\dim(H_0) < \dim(K_0) = \dim(K)$. In the former case we clearly have $[H:H_0] < [K:H_0] = [K:K_0]$. Hence in both cases we have $|H| < |K|$. $\square$
So other than “are there any mistakes I don't know about?” I'm asking “if $H$ is a closed subgroup of the algebraic group $K$, is it always the case that $H_0 \leq K_0$?”
Background: My setup for algebraic groups is groups whose underlying set is a subset of the set of invertible matrices over an algebraically closed field that is the zero set of a set of polynomials in its entries and determinant, such that multiplication and inversion are similarly given by polynomials in the entries and determinant. I think that's a standard definition, so feel free to fix it if I stated something incorrectly.
If you are flexible on references, I prefer the textbook by Malle–Testerman (2011).
Motivation: I want to show that unipotent groups have a useful normalizer condition, so I want to say they have ACC on closed subgroups. However, I noticed that $K^\times$ lacks ACC on closed subgroups, and by extension every torus, and by extension every Borel subgroup. However, I couldn't think of any violations of DCC, and I read about this "dimension" idea, and it seemed plausible that I could give this analogue of order. My hope is that ACC for closed unipotent subgroups will hold for similar dimension reasons, but first thing is first.