"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

(1)

Suppose

(2)

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

we know that for any i

(3)

is an irrc chain in U_i

i.e.

(∀i)dimX	≤ dimUi
dimX	≤ supdimUi

So (1) has been shown.

Now of course we show the reverse inequality:

(4)

Suppose now, given i,

(5)

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

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

(6)

is an irrc chain in X
Hence...

(∀i)dimUi	≤ dimX
supdimUi	≤ dimX

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

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


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