← I.1.10a I.1.10d →



"I'M BACK", the fool says, only to drop out for another week. THE FOOL. SOMEONE, GET THE GUN. WALLS. SPLATTER. PINK. Etc.

Ugh. Well, this exercise is pretty simple. I mean... it's more of a PART of an exercise than an entire exercise. But, y'know, me. Hey, you baked a pie? Lovely! What flavor is your pie? Pumpkin? Apple? PFFFT. Like I care. I'm stuck with a single broken off wedge from the edge of its fluted crust. So here's the theme song for today's thread. Yes the edges of a pie crust are neatly, delicately, symmetrically crimped by feminine hands, while the subject of that fugue is an incongruently twisting gorge culiminating in a reckless trill. We are, after all, in Hartshorned, not a cooking class. "B-but the beauty of math is in symmetry and perfection". *Vomits on your face*.

And thus, let's commence: We'll start by showing

dim X ≤ supdim Ui


C1 ⊊ ⋅⋅⋅ ⊊ Cn

is an irrc chain in X. With some help from an old friend:

we know that for any i

Ui ∩ C1 ⊊ ⋅⋅⋅ ⊊ Ui ∩ Cn

is an irrc chain in Ui


(i)dimX dimUi
=⇒dimX supdimUi

So (1) has been shown.

Now of course we show the reverse inequality:
sup dim Ui ≤ dim X

Suppose now, given i,

C1 ⊊ ⋅⋅⋅ ⊊ Cn

is an irrc chain in Ui
Now use the SECOND sentence from our old friend:

i.e. we know that, taking closures in X,

clX(C1) ⊊ ⋅⋅⋅ ⊊ clX (Cn)

is an irrc chain in X

(i)dimUi dimX
=⇒supdimUi dimX

Explanation: The first inequality says that dimX is an upper bound on dimUi. Since supdimUi is the LEAST upper bound on dimUi, we must have supdimUi dimX.

And thus, (4) has been shown. CoOOoOOoOOOooOOOoOOOOOOooOOl! (See the wedges?)

Holy, fuck, I actually typed "So here's the theme song for today's thread."