Recent content by amanda_ou812

Question: Let T be a diagonalizable linear operator on a finite-dimensional vector space, and let m be any positive integer. Prove that T and T^m are simultaneously diagonalizable. Definition: Two linear operators T and U are finite-dimensional vector space V are called simultaneously...

so, is Sn-Sn+1 supposed to equal 1/n - 1/(n+1)? So, that you then get Sn-Sn+1 = to 1/(n(n+1))?

Pathwise connected

do I do it peicewise? like f(p) = (x,y) for p=0, (u, v) for p=1 and then this other functions that I cannot figure out how to get.

i want to say something like given f(0) = (x, y) and p in [0,1] then f(p) = ((y-v)/(x-u)) * p + (x, y) but that's not correct bc i am switching between R and R^2 do i need to make a continuous function for each component?

i am not sure what you mean by function language but I would want f(0) = (x,y) and f(1) = (u, v)

if I take a straight line path from the points, it would be the distance between the two, so sqrt((x-u)^2 + (y-v)^2)

I need to prove that X={(x,y):a<=x<=b, c<=y<=d} I was thinking of using proof by contradiction. Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two. I can show that then the set is disconnected but not sure where to go after...

So, since there exists a t contained in R so that d(t, f(K)) = inf { |t-s|: s in contained in f(K)} there is a number km such that for any k contained in K, f(km) = inf{f(K) : k in K} and f(km) <= f(k). But, f(km) = d(t, f(K))...I don't think I am understanding the problem. So, I defined a...

I used x-0 to mean x-nought as an arbitrary point in X. If I use x and y's doesn't that prove its uniformly continuous? (of course, that would mean continuous as well) f(K) compact implies f(K) is a closed and bounded subset of R which implies f(K) has a sup and inf. When p-nought is...

Ah, I see your point. Given e>0 let delta = e the for x-0 contained in X we have d(x, x-0)< delta implies abs (f(x) - f(x-0)) = abs ( d(p, x) - d(p, x-0)) <= abs (d(x, x-0)) [reverse triangle inequality] < delta = e. Fredrik - f(K) is compact

HELP! real analysis question: continuity and compactness Homework Statement Let (X,d) be a metric space, fix p ∈ X and define f : X → R by f (x) = d(p, x). Prove that f is continuous. Use this fact to give another proof of Proposition 1.126. Proposition 1.126. Let (X, d) be a metric space...

Homework Statement Let (X,d) and (Y, p) be metric spaces and f : X -> Y a function. Prove that f is continuous at p-0 if and only if for every ε > 0, there exists a δ > 0 so that the image of Bd(p-0; δ) is contained in Bp( f (p-0);ε). Homework Equations The Attempt at a Solution...

So, how did you choose those numbers? (1-r/2) and (1+r/2). I would like to know so that can learn how to do this type of problem. And, I show that (1-r/2)P is in B(p, r) by showing that d((1-r/2)p, y)<r and I show that it is in S by showing that (1-r/2)P and y satisfies the condition of S...

hm...I see your point. So, {pn} infinite and {pn} as a subset of B(p; r) intersection S does not imply that B(p;r) intersection S has indefinitly many points. Is there another way I can get to that conclusion?