← I.1.10b I.1.12 →



I don't know how to do part (c) yet, so I'm doing part (d).

Right off the bat we know Y X.

Let dim X = n. Since we know that dim Y = n, there is an irrc chain in Y :

C1  ⊊ ⋅⋅⋅ ⊊ Cn

Note that the above is also an irrc chain in X. Now, suppose Y X. Then Cn Y = ⇒Cn X. And by virtue of X being an irrc subset of itself,

C  ⊊  ⋅⋅⋅ ⊊ C  ⊊ X
 1           n

is an irrc chain in X. Which would mean dim X n + 1, a contradiction.
Hence Y = X.
Fuck that was clever. Too clever by my standards. At least I have part (c) to remind me of my incompetence. You'll see!

I received many complaints (JK, obviously, no one reads this. Though from that perspective, "many complaints" is technically correct. "The equations are too small!" someone says. But who is this someone, the textual voice jutting in with quotes? Is it you, dear reader? Or is it me? A book is meant to be a dialogue between the author and the reader. But an obscure personal website? That's between an admin and himself. Nay: It ATTEMPTS to be a conversation between oneself and someone, just anyone else out there, but it ENDS UP turning into selfcest... wut?) that the math was too small, so I used \DisplayMath to increase the size.