After Midterm Stuff

Front 1

Continuity

Back 1

iff for every E>0, there exists a d>0 such that whenever xinD with |x-x0|<d we have |f(x)-f(x0)|<E

Front 2

Theorem 19: let f:dR be a function x0inD and x0 is an accu point of D. TFAE
1) f is cont at x0
2)f has a limit at x0 and limf(x)=f(x0)
3)whenever (xn) in D converging to x0, the sequence f(xn) converges to f(x0)
3=>2

Back 2

3=>2
Suppose 3 holds.let xn be a seq in D\{x0} which converges to xo. then by 3 f(xn) converges to f(x0). then by thrm 16 f has a lim at xo, namely f(x0)

Front 3

Theorem 19: let f:dR be a function x0inD and x0 is an accu point of D. TFAE
1) f is cont at x0
2)f has a limit at x0 and limf(x)=f(x0)
3)whenever (xn) in D converging to x0, the sequence f(xn) converges to f(x0)


2=>1

Back 3

suppose 2 holds. then there exists an E>0 and d>0 such that whenever xinD\{x0} with |x-x0|<d we have |f(x)-f(xo)|<E. note that when x=x0, we have |f(x0)-f(x0)|=0<E. thus for all xinD with |x-x0|<d we have |f(x)-f(x0)<E. thus f is continous

Front 4

Theorem 19: let f:dR be a function x0inD and x0 is an accu point of D. TFAE
1) f is cont at x0
2)f has a limit at x0 and limf(x)=f(x0)
3)whenever (xn) in D converging to x0, the sequence f(xn) converges to f(x0)

1=>3

Back 4

suppose 1 holds. let xn be any sequence in D converging to x0. let E>0 be given. by 1 there exists a d>0 s.t when xinD x/ |x-x0|<d we have |f(x)-f(x0)|<E . since limnx=x0 we may pick an N s.t for all n with n≥N we have |f9xn)-f(x0)|<E thus f(xn) converges to f(x0)

Front 5

if f:D-->R is a function DcR, x0inD and x0 NOT an accu point of D. then f is continuous

Back 5

Let E>0 be given. that xo is not an accu point implies by a thrm that there exists a neighborhood of x0 s.t QINTDc{x0}. pick d>0 s.t (x0-d, x0+d)cQ and let xinD w/
|x-x0|<d. by our choice of d, this implies x in QINTD and hence x=x0. Consequently |f(x)-f(x0)|=|f(x0)-f(x0)|=0<E

Front 6

if f:D-->R is a function DcR, xinD and x0 NOT an accu point of D. then f is continuous

Back 6

Let E>0 be given. that xo is not an accu point implies by a thrm that there exists a neighborhood of x0 s.t QINTDc{x0}. pick d>0 s.t (x0-d, x0+d)cQ and let xinD w/
|x-x0|<d. by our choice of d, this implies x in QINTD and hence x=x0. Consequently |f(x)-f(x0)|=|f(x0)-f(x0)|=0<E

Front 7

if f:D-->R is a function DcR, xinD and x0 NOT an accu point of D. then f is continuous

Back 7

Let E>0 be given. that xo is not an accu point implies by a thrm that there exists a neighborhood of x0 s.t QINTDc{x0}. pick d>0 s.t (x0-d, x0+d)cQ and let xinD w/
|x-x0|<d. by our choice of d, this implies x in QINTD and hence x=x0. Consequently |f(x)-f(x0)|=|f(x0)-f(x0)|=0<E

Front 8

Theorem 20: let f,g:DR be function x0inD suppose that both f,g are both cont at x0 then

2) fg is cont at x0

Back 8

from arithmetic of lims we know limflimg=limfg=f(xo)g(xo). thus by thrm 19 fg is cont

Front 9

Theorem 20: let f,g:DR be function x0inD suppose that both f,g are both cont at x0 then
3) if g(x0)≠0 f/g:D’{xinD|g(x0)≠0}R x|f(x)/g(x)

Back 9

by 2 it suffices to show tprove that for g(x0)≠0 the finction 1/g:D’-->R is cont at x0. Let E>0 and d1>0 such that whenever |x-x0|<d1 xin D’, then |g(x)|≥M. since g is cont at x0 there exists a d2>0 s.t for all x in D’ w/ |x-x0|<d2 we have |g(x)-g(x0)|<E*M*|g(x0)|, using the fact that E*M*|g(x0)|>0. Let d=min(d1,d2), then d>0. Then for all x in D’ w/ |x-x0|<d we conclude
|1/g|=|g(x)-g(x0)| =
|g(x)-g(x0)|/|g(x)||g(x0)| < EM|g(x0)/M|g(x0)| =E

Front 10

Let g:D-->R be a function x0inD. And g(x0)≠0. Then there exists M>0 and d>0 s.t for all xinD with |x-x0|<d we have |g(x)|>=M

AKA?

Back 10

Let g:D-->R be a function x0inD and g(x0)not=0. then there exists M>0 and a nbhd Q of xo such that whenever xinQINTD we have |g(x)|>=M

Front 11

Let g:D-->R be a continous function x0inD. And g(x0)≠0. Then there exists M>0 and d>0 s.t for all xinD with |x-x0|<d we have |g(x)|>=M

Back 11

Let M=|g(x0)|/2. since g is cont there exists a E=M and d>0 s.t whenever xinD with |x-x0|<d we have |g(x)-g(x0)|<E. then |g(x)|=|g(x)-g(x0)+g(x0)|>=||g(x)-g(xo)|-|g(x)||=||g(x0)|-|g(x)-g(xo)||<2M-M=M

Front 12

closed set

Back 12

S is closed if S contains all its accu points

Front 13

open set

Back 13

S is open if for each xinS there exists a neighborhood Q of x s.t QcS

Front 14

Let ScR. Then S is closed(open) iff R\S is open (closed)-(corollary)
=>

Back 14

=> suppose S is closed. let xinR/S. thus xnotinS and therefore isnt an accu point of S since S is closed. thus there exists a nbhd of x such that QINTSc{x}. this implies xinS, but this cant be true so QINTS=emptyset, i.e QcR\S. thus R\S is open.

Front 15

Let ScR. Then S is closed(open) iff R\S is open (closed)-(corollary)
<=

Back 15

<= assume R\S is open. Let x0 be an accu point of S. If x0 in S, good. Suppose x0 ≠inS. Then x0inR\S. since R\S is open, there exists a neighborhood Q of x0 so that QcR\S. but QᴒS is empty so x0 could not be an accu point. This is a contradiction, thus x0inS for all accu pnts, so S is closed

Front 16

Theorem 24 (Balzano): Let f:[a,b]-->R be a continuous function. Let a,binR w/ a<b and suppose that f(a) and f(b) have opposite signs. Then there exists c in(a,b) w/ f(c)=0

Back 16

suppose f(a)<0 and f(b)>0 (other case is proved symmetrically).we will by induction on nconstruct 2 sequences (xn) and (yn) in [a,b] such that for each n the following conditions are satisfied
1)f(xn)<0 and f(yn))≥0
2)a=x1x1≤x2≤. . .≤n<yn≤yn-1≤. . . ≤y1=b
3)y¬i-1-xi-1=(1/2i)(b-a) for i=1, . . ,n
Construction of xn and yn. by induction on (n=1): x1=a, y1=b. (induction step) suppose we have constructed x1. . .xn,y1. . . ,yn such that 1-3 are satisfied. define Cn=xn+(yn-xn)/2. if f(cn)<0 define xn+1=xn and yn+1=Cn.
then (xi)(iinN+1) and
(yi)(iinN) would again satisfies 1-3. Note 2,3 are satisfied case independently, and 1 is satisfied comes from the induction hypothesis. thus all parts of the induction hypothesis are re-established for (xi)i≤n+1 and (yi)n+1. hence induction yields infinite sequence xn and yn satisfies 1-3. now by 2 xn is increasing and bounded from above and thus converges. analogously yn converges. by 3 sequences (yn-xn) converges to 0. and hence by arithmetic of lims we get limxn=limyn=c. note cin[a,b]. finally we show that f(c)=0.since limxn=limyn=c and f is cont at c, thrm 21 yields limf(xn)=lim(yn))≠f(c). now by 1 limf(xn)≤0 and limf(yn)≥0, hence f(c)=0.

Front 17

Let f:[a,b]-->R be a continuous function w/ f(a)≠f(b). Moreover suppose z lies strictly between f(a) and f(b). Then there exists cin(a,b) with f(c)=z

Back 17

suppose f(a)<z<f(b) (the other case is proved symmetrically). let g:[a,b]-->R with g(x)=f(x)-z. then g(x) is also cont. apply thrm 24 to g, we obtaim g(c)=0=f(c)-z, so f(c)=z.

Front 18

Theorem 23: Lat f:RR, then TFAE
1. F is continous
2. For each open set UcR, the inverse image f-1(U) is open
1=>2

Back 18

Suppose 1 holds. let x0inf-1(U). since U is open there exists a nbhd of Q of f(x0) so that QcU. then there exists E>0 so that (f(x0)-E, f(x0)+E)cQ. since f is cont at x0, there exists d>0 so that for all x in R w/ |x-x0|<d we have |f(x)-f(x0)|<E. set q':=(xo-d,x0+d) this a nbhd of x0 for xinQ' we have |x-x0|<d. so |f(x)-f(x0)|<E. now f(x) in f(x0)-E,f(x0)+E)cQcU. So xinf-1(U). therefore f-1(U) is open since Q'cf-1(U)

Front 19

Theorem 23: Lat f:RR, then TFAE
1. F is continous
2. For each open set UcR, the inverse image f-1(U) is open
2=>1

Back 19

assume the inverse image under f of every open set is open. let E>0, let U=(f(x)-E,f(x)+E) this is an open set so f-1(U) is open by hypothesis. Note: since f-1(U) is open there exists a neighborhood Q of x0 s.t Qcf-1(U) therefore there exists d>0 s.t (x0-d,x0+d)c for any xinR w/ |x-x0|<d, we have
x in (x0-d, x0+d)cQcf-1(U), so f(x)inU and thus
|f(x)-f(x0)|<E. thus f is cont at x0. thus f is cont