130B Test 1 Flashcards
A set A as a subset of reals is compact iff
Every open cover of A has a finite subcover.
A function f:D-Reals is uniformly continuous on E (subset of D) iff
For every €>0 given, there exists a d>0 such that if x,y€E and
|x-y|<€
If f is uniformly continuous on D, then f is uniformly continuous.
What does it mean that a set E (subset of R) is closed?
E contains all of its accumulation points.
What does it mean that a set E (subset of R) is open?
For each x€E, there exists a neighborhood Q of x such that Q is a subset of E.
A set E is compact iff
For every family {Ga}a€A of open sets such that E is a subset of
U(a€A)Ga, there exists a finite set {a1,…,an} as a subset of A such that E is a subset of U(i=1to n)Gai
•In other words, E is compact iff every open cover of E has a fnite subcover.
Heine-Borel Theorem
A set E(subset of R) is compact iff E is closed and bounded.
Let f:D-R be continuous with D compact (closed and bounded), then f is
Uniformly continuous
If f:D-R is uniformly continuous and D is a bounded set, then
f(E) is a bounded set.
Let f:E-R be continuous with E compact, then
f(E) is compact.
Bolzano Intermediate Value Theorem: Let f:[a,b]-R be continuous with f(a)<f(a)], then
There exists a c€(a,b) such that f(c)=y
If f:[a,b]-R be continuous and 1-1, then
F is monotone
Let x. be an accumulation point of a set D(subset of R) and x.€D, then f is differentiable at x. iff
•For x€D{x.}
•the limit(x->x.) of T(x)=[f(x)-f(x.)]/(x-x.) exists. If this limit exists,
then it is the derivative of f at x..
•the limit(t->0) of [f(x+t)-f(x.)]/t exists. If this limit exists, then it is
the derivative of f at x..
•For every sequence {Xn}(n=1->infinity) of points of D{x.} converging to x., the sequence {[f(Xn)-f(X.)]/(Xn-X.)}(n=1->infinity) converges.
Let f:D->R be differentiable at x., then
x.€D, x. Is an accumulation point of D, and f is continuous at x..
Let f,g:D->R be differentiable at x.. What can you say about f+g, fg, and f/g?
- (f+g)’(x.)=f’(x.)+g’(x.) exists
- (fg)’(x.)=f(x.)g’(x.)+f’(x.)g(x.) exists
- (f/g)’(x.)=[g(x.)f’(x.)-f(x.)g’(x.)]/[g(x.)]^2 exists given that g(x.) doesn’t = 0
Chain rule: If f:D->R and g:D’->R with f(D) a subset of D’
If f is differentiable at x. and g is differentiable at f(x.), then what can be said about g•f (g composed of f)?
(g•f)’(x.)=g’(f(x.))f’(x.) exists