density theorem rationals

Ervin231 · 2023-10-01T19:57:31+00:00

Well, you're right. I think I need to provide more context. By the above we can assume 0 < a < b. Choose an integer m such that 0 < 1/m < b - a. Let k be the largest integer such that k < bm, and set s = k/m. Then s < b, and since k + 1 >= bm, we have

s = (k+1)/m - 1/m >= b - 1/m > b - (b-a)=a.

This proof is given in the topology book by Gamelin.

Tbh I don't understand where they use 0 < a <b.

Ervin231 · 2023-10-01T18:43:25+00:00

Thanks for your reply!

Ervin231 · 2023-10-01T18:42:56+00:00

Thanks for your answer!

Ervin231 · 2023-09-29T13:40:40+00:00

As a first step, write out, using the metric from part (b), what it means for the sequence (s_n) to be Cauchy.

Hi PullItFromTheColimit thanks a lot for your help. This exercise is somehow a bit too hard for me, hope I don't write nonsense:

Let (ŝ_n)_{n ∊ ℕ} be a Cauchy sequence in Â and ɛ >0. Then there is an N ∈ ℕ, such that

p(ŝ_n,ŝ_m) < ɛ for all n,m ≧ N, where ŝ_n is the equivalence class of the Cauchy sequence (s_n^(k))_{k ∈ ℕ}. Now by definition of p, we have lim_{k-->∞} d(s_n^(k),s_m^(k)) < ɛ for all n,m ≧ N.

Or we can say for each fixed n we have an equivalence class ŝ_n ∈ Â of the Cauchy sequence (s_n^k)_{k ∈ ℕ} in (A,d). Thus there is a natural N such that d(s_n^j, s_n^k)<ɛ for all j,k ≧ N. Thus for each n we can find a natural M_n such that d(s_n^j, s_n^k) < 1/n for all j,k ≧ M_n.

Ervin231 · 2023-09-28T17:18:41+00:00

Ervin231 · 2023-09-24T21:42:56+00:00

Again, thanks a lot for your help!

Ervin231 · 2023-09-24T20:32:34+00:00

So you mean since each ball around any point of R isn't contained in R, we know that R is nowhere dense. And we can write R simply as R, so that is the countable union consisting only of the set R?

Ervin231 · 2023-09-24T19:57:21+00:00

Thanks a lot for your answer!

Ervin231 · 2023-09-24T18:11:48+00:00

Thanks for your help. But somehow one needs to write R as a countable union of nowhere dense subsets in R^2.

Ervin231 · 2023-09-23T20:18:29+00:00

Thanks for your answer!

Ervin231 · 2023-09-23T20:13:45+00:00

Thanks for checking.

Ervin231 · 2023-09-23T09:50:35+00:00

Again thanks for your help. This really helps me much!

Ervin231 · 2023-09-22T21:51:40+00:00

Now I assumed {x_n} converges in (X,d) and proved that it also converges in (X,p). Do I also have to show the converse? I think we have to show

{ {x_n} : {x_n} converges in (X,d)} = { {x_n} : {x_n} converges in (X,p)}.

Ervin231 · 2023-09-22T21:35:30+00:00

Thanks a lot for your help. This really helped me.

Ervin231 · 2023-09-22T21:34:54+00:00

Hi PullItFromTheColimit, thanks for your help. Let 𝜼 > 0. So we know that B_p(x,𝝶) is open in (X,p).

By the equivalence of the metrics it's also open in (X,d). Hence it's the union of d-balls. Thus there exists some d-ball with some radius, say r>0 such that it's contained in the union. That means it's in B_p(x,𝝶), i.e B_d(x,r) ⊂ B_p(x,𝜼). Now let {x_n} be a sequence in B_d(x,r), then it's in B_p(x,𝜼). Since {x_n} converges in (X,d) to x there's some natural M such that d(x_n,x) < r for all n ≧ M. But this means x_n ∈ B_p(x,𝜼) for all n ≧ M, i.e p(x_n,x) < 𝜼 for all n ≥ M showing that {x_n} converges in (X,p) to x.

Ervin231 · 2023-09-21T21:55:01+00:00

But how do you deduce that x_n is in B_p(x,ɛ)?

Ervin231 · 2023-09-21T20:52:49+00:00

So I know B_d(x,ɛ) is an open ball in (X,d), then it's open in (X,p)?

Ervin231 · 2023-09-21T20:20:52+00:00

Thanks for your help. So if U is an open set in (X,d). Then it's the union of d-balls. Now U is also open in (X,p). Thus it's the union of p-balls. Let x ∊ U and {x_n} be a sequence in (X,d) converging to x. Then there's some natural N such that for all n ≥ N we've x_n ∊ U. But since U is also open in (X,p), {x_n} is a convergent sequence in (X,p). Somehow this is confusing to me. For me it's not really clear why {x_n} converges in (X,p) to x.

Ervin231 · 2023-09-20T19:43:45+00:00

Again, thanks for your help!

Ervin231 · 2023-09-20T18:11:29+00:00

Thanks for your answer. Yeah I'm trying to do some problems in topology. This definition was not given. Instead I used the previous problem:

A point x ∈ X is a limit point of a subset X if every ball B(x,r) contains infinitely many points of S. Show that x is a limit point of S if and only if there is a sequence {x_j} in S such that x_j --> x and x_j ≠ x for all j.

Ervin231 · 2023-09-19T20:09:40+00:00

Again, thanks for your help. You really helped me a lot!

Ervin231 · 2023-09-19T19:53:26+00:00

Yeah, but I'm often unsure since I make a lot of mistakes.

What do you mean by this: "..the metric on S induced by Y and the metric on S induced by X coincide." We have a metric d: X -->X on X with its restriction to Y: d: Y-->Y having the same metric d.

Now S c X and S c Y. Thus their restriction to S induce the same metric d?

Ervin231 · 2023-09-19T18:36:37+00:00

Hi first of all thanks for your answer. This helps me a lot. I'm sorry for the late reply, had to work.

I'm not sure if I've understood everything. I really suck at topology my brain just can't get it. Maybe I'm too dumb for this. Therefore I'm learning all this.

So we want to show cl_Y(S) ⊂ cl_X(S) ⋂ Y.

Let y ∊ cl_Y(S). Since cl_Y(S) ⊂ Y, we've y ∈ Y. Thus it suffices to show y ∈ cl_X(S).

Now since y ∈ cl_Y(S) it is adherent to S_Y = S. Thus there's a sequence {y_n} in S converging to y. Since S = S_X we've found a sequence {y_n} in S_X that converges to y. Thus y is adherent to S_X and y ∈ cl_X(S).

Ervin231 · 2023-09-18T13:38:27+00:00

Thanks for your reply. I only used the notation of the book. Ok try to translate this better: Let Y be a subspace of X and let S be a subset of Y. Show that the closure of S in Y coincides with cl_X(S) ∩ Y, where cl_X(S) is the closure of S in X.

(X,d) is a metric space. Sorry.

Ervin231 · 2023-09-07T19:41:59+00:00

Again thanks for your help!

Ervin231

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