Metric Spaces UPDATED ACTUAL Exam Questions and CORRECT Answers

Metric Spaces UPDATED ACTUAL Exam
Questions and CORRECT Answers
metric space - CORRECT ANSWER - set with a distance function that must be positive,
symmetric and obey the triangle inequality


generalisation of the triangle inequality - CORRECT ANSWER - more than two basically,
prove with transitivity and induction


l infinity space - CORRECT ANSWER - the set of all bounded sequences of real numbers.
{x={xn}|xn in R s.t. for all n in N there exists Mx s.t. |xn|<Mx}. With metric d(x,y)=sup|yn-xn|
which exists because y and x are bounded


C[a,b] as a metric space - CORRECT ANSWER - {f:[a,b]->R|f is cts} with metric
d(f,g)=max|f(x)-g(x)|


alternative C[a,b] metric - CORRECT ANSWER - integral of |f-g|



discrete metric spaces - CORRECT ANSWER - any set whatsoever with metric d(x,y)=0 if
x=y and 1 if x=/=y


lp spaces - CORRECT ANSWER - {x={xn}|E|xi|^p<+inf} R or C. with metric
d(x,y)=(E|xi-yi|^p)^1/p


young's inequality - CORRECT ANSWER - ab<=a^p/p+b^q/q where a,b>0 p>1 and q is
the conjugate exponent of p


conjugate exponent of p - CORRECT ANSWER - q such that 1/p+1/q=1

,proof of young's inequality - CORRECT ANSWER - take the graphs of different cases and
then use integration. (the graph of y=x^p-1, a on x, b on y)


holder's inequality - CORRECT ANSWER - E|xiyi|<=(E|xi|^p)^1/p(E|xi|^q)^1/q where x
in lp and y in lq


proof of holder's inequality - CORRECT ANSWER - define x_ and y_ like
xi_=xi/(E|xi|^p)^1/p. Then they're both in lp and lq still. In actual fact E|xi_|^p=1 and so use
young's inequality with a=|xi_| and b=|yi_| to get that E|xi_yi_|<=1 then sub in to finish


minkowski's inequality - CORRECT ANSWER -
(E|xi+yi|^p)^1/p<=(E|xi|^p)^1/p+(E|yi|^p)^1/p


proof of minkowski's inequality - CORRECT ANSWER -



open balls, closed balls, spheres - CORRECT ANSWER - B(x,r)={y in X: d(x,y)<r}.
B~(x,r)={y in X: d(x,y)<=r}. S(x,r)={y in X: d(x,y)=r}. BUS=B~


balls and spheres in function spaces - CORRECT ANSWER - key concept is ribbons.
Sphere must touch edge


open set - CORRECT ANSWER - all points are interior points



closed set - CORRECT ANSWER - complement is open (has boundary)



open balls are open sets proof - CORRECT ANSWER - use triangle inequality



closed balls are closed sets proof - CORRECT ANSWER - open balls are open sets and
contradiction by triangle inequality

, basic definition of a topology U - CORRECT ANSWER - like a power set but with the
extra condition that every set in U is defined to be an open set


axioms of a topology U of X - CORRECT ANSWER - empty set and X are in U, if Ei is in
U then the intersect from 1 to n of Ei is in U, if Ea is in U then the union over a of Ea is in U


topological space - CORRECT ANSWER - (X,U)



relationship between metrics and topology - CORRECT ANSWER - metrics give rise to a
topology but not necessarily the other way


hausdorff topological space - CORRECT ANSWER - if the topology specifies a metric



closed set, limit points - CORRECT ANSWER - a closed set contains all its limit points



metrical def of continuity - CORRECT ANSWER - for all e>0 there exists d(x,e)>0 st if y
in B(x,d(x,e)) then f(y) in B(f(x),e)


topological def if continuity - CORRECT ANSWER - if the preimage of an open set is an
open set


proof that metrical continuity implies topological continuity - CORRECT ANSWER -
open set, interior, continuity, containment, open set, top


proof that topological continuity implies metrical continuity - CORRECT ANSWER -
open set in codomain, open set in domain, interior point, got ya d


pseudometric spaces - CORRECT ANSWER - don't have to have positive definiteness, i.e.
d=0 doesn't have to mean x=y