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### 19 Cards in this Set

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 Continuity iff for every E>0, there exists a d>0 such that whenever xinD with |x-x0|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) 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 suppose 2 holds. then there exists an E>0 and d>0 such that whenever xinD\{x0} with |x-x0|3 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|R is a function DcR, x0inD and x0 NOT an accu point of D. then f is continuous 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|R is a function DcR, xinD and x0 NOT an accu point of D. then f is continuous 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|R is a function DcR, xinD and x0 NOT an accu point of D. then f is continuous 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|R is cont at x0. Let E>0 and d1>0 such that whenever |x-x0|0 s.t for all x in D’ w/ |x-x0|0. Let d=min(d1,d2), then d>0. Then for all x in D’ w/ |x-x0|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|=M AKA? 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 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|=M Let M=|g(x0)|/2. since g is cont there exists a E=M and d>0 s.t whenever xinD with |x-x0|=||g(x)-g(xo)|-|g(x)||=||g(x0)|-|g(x)-g(xo)||<2M-M=M closed set S is closed if S contains all its accu points open set S is open if for each xinS there exists a neighborhood Q of x s.t QcS Let ScR. Then S is closed(open) iff R\S is open (closed)-(corollary) => => 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. Let ScR. Then S is closed(open) iff R\S is open (closed)-(corollary) <= <= 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 Theorem 24 (Balzano): Let f:[a,b]-->R be a continuous function. Let a,binR w/ a0 (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≤. . .≤nR 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 suppose f(a)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. 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 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|1 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|