Advanced calculus, prove the bijection

NOTES WEEK 02 DAY 2

SCOT ADAMS

Gave hints for HW#2 and HW#3.

Reviewed 99K, 99Ką, Ă99K, Ñ, etc. Reviewed picturing f : X Ă99K Y in two ways.

DEFINITION 0.1. Let X and Y be sets. Then DX Ă99K Y means: Df Ď X ˆ Y such that f : X Ă99K Y . Also, DX 99Ką Y means: Df Ď X ˆ Y such that f : X 99Ką Y . Also, DX Ă99Ką Y means: Df Ď X ˆ Y such that f : X Ă99Ką Y .

For example, Dt1, 2u Ă99K t3, 4, 5u, but Et3, 4, 5u Ă99K t1, 2u. Also, Dt3, 4, 5u 99Ką t1, 2u, but Et1, 2u 99Ką t3, 4, 5u.

Notational convention. We often want to describe functions with an

infinite domain, like, say, tpx, x2q |x P Ru : R Ñ R. However, writing, say, “Let f :“ tpx, x2q |x P Ru : R Ñ R” is highly nonstandard. Instead, we will write “Let f : R Ñ R be defined by fpxq “ x2” or “Define f : R Ñ R by fpxq “ x2”. Note that, in the last sentence, x is a free variable. The definitions of functions are the one place in this

course where we will tolerate free variables. It would be more correct

to write “Let f : R Ñ R be defined by: @x P R, fpxq “ x2”. However, this is also highly nonstandard, and the convention is that we skip the

quantification here.

FACT 0.2. DN ĂÑą N0.

Proof. Define f : N Ñ N0 by fpkq “ k ´ 1. Then f : N ĂÑą N0. Then DN ĂÑą N0. �

DEFINITION 0.3. For any set S, define idS : S Ñ S by idSpxq “ x.

FACT 0.4. For any set S, DS Ă99Ką S.

Proof. Unassigned HW. Hint: Use idS. �

Date: September 15, 2016 Printout date: September 16, 2016. 1

2 SCOT ADAMS

DEFINITION 0.5. Let A, B, C be sets. Let f : A 99K B. Then

f˚pCq :“ t fpxq | x P C Xpdomrfsq u and

f ˚pCq :“ t x P domrfs | fpxq P C u.

We pictured f˚pCq and f ˚pCq.

We will often write fpCq to mean f˚pCq.

Some people use f ´1pCq to mean f ˚pCq, but we will avoid this nota-

tion, for fear of confusion with the inverse function f ´1 defined below.

DEFINITION 0.6. Let A and B be sets. Let f : A ĂÑą B. Then

f ´1 : B Ñ A is defined by f ´1pyq “ ELTrf ˚ptyuqs

We pictured f ´1. We call f ´1 the inverse of f.

E.g.:If f “ tp1, 2q,p2, 3q,p3, 4qu, then f ´1 “ tp2, 1q,p3, 2q,p4, 3qu.

FACT 0.7. Let A, B be sets, f : A ĂÑą B. Then f ´1 : B ĂÑą A.

Proof. Unassigned homework. �

FACT 0.8. For any sets S, T , we have:

pDS Ă99Ką Tq ñ pDT Ă99Ką Sq

Proof. Unassigned HW. Hint: Use inverse functions. �

DEFINITION 0.9. Let A, B, C, D be sets. Let f : A Ñ B and let

g : C Ñ D. Then g˝f : f ˚pCq Ñ D is defined by pg˝fqpxq “ gpfpxqq.

Assigned HW#7.

LEMMA 0.10. Let X, Y, Z be sets. Let f : X ĂÑ Y , g : Y ĂÑ Z.

Then g ˝ f : X ĂÑ Z.

Proof. We have domrg ˝ fs “ f ˚pdomrgsq “ f ˚pY q “ X. It remains to

prove: @a, b P X, rpg ˝ fqpaq “ pg ˝ fqpbqs ñ ra “ bs.

We know: @s, t P X, rfpsq “ fptqs ñ rs “ ts.

We know: @s, t P Y , rgpsq “ gptqs ñ rs “ ts.

Given a, b P X. We wish to prove: rpg˝fqpaq “ pg˝fqpbqs ñ ra “ bs.

Assume pg ˝ fqpaq “ pg ˝ fqpbq. We wish to prove a “ b.

We have gpfpaqq “ gpfpbqq. Then fpaq “ fpbq. Then a “ b. �

Assigned HW#8.

Recall that R˚ :“ R Yt8,´8u.

NOTES WEEK 02 DAY 2 3

DEFINITION 0.11. Let S Ď R˚ and let a P R˚. Then S ď a means: @x P S, x ď a.

Also, S ě a means: @x P S, x ě a.

Also, S ă a means: @x P S, x ă a.

Also, S ą a means: @x P S, x ą a.

Also, a ě S means: @x P S, a ě x.

Also, a ď S means: @x P S, a ď x.

Also, a ą S means: @x P S, a ą x.

Also, a ă S means: @x P S, a ă x.

“S ď a” and “a ě S” are equivalent. They are read “S is bounded

above by a” and “a is an upper bound for S”.

“S ą a” and “a ă S” are equivalent. They are read “S is strictly

bounded below by a” and “a is a strict lower bound for S”.

DEFINITION 0.12. Let S Ď R˚. Then UBpSq :“ ta P R˚ |a ě Su and LBpSq :“ ta P R˚ |a ď Su.

That is UBpSq is the set of upper bounds for S and LBpSq is the set

of lower bounds for S.

DEFINITION 0.13. Let S Ď R˚. Then max S :“ ELTpSXrUBpSqsq and min S :“ ELTpS XrLBpSqsq.

We calculated maxt1, 2, 3u “ 3.

We calculated maxr1, 2q “ /. We calculated minr2,8s “ 2.

We calculated max Z “ / “ min Z. We calculated max N “ / and min N “ 1. We calculated max N0 “ / and min N0 “ 0. We calculated max H “ / “ min H.

DEFINITION 0.14. Let S Ď R˚. Then sup S :“ minpUBpSqq and inf S :“ maxpLBpSqq.

Sometimes sup S is called the “least upper bound of S”.

Sometimes inf S is called the “greatest lower bound of S”.

We calculated supt1, 2, 3u “ 3.

We calculated supr1, 2q “ 2.

We calculated infr2,8s “ 2.

We calculated sup Z “ 8 and inf Z “ ´8.

4 SCOT ADAMS

We calculated sup N “ 8 and inf N “ 1. We calculated sup N0 “ 8 and inf N0 “ 0. We calculated sup H “ ´8 and inf H “ 8.