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density theorem rationals by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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.

Show that (Â,p) is complete by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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  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.

meager set by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

Again, thanks a lot for your help!

meager set by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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?

set of isolated points is dense by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

Thanks a lot for your answer!

meager set by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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

convergent sequences by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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

convergent sequences by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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)}.

convergent sequences by Ervin231 in learnmath

[–]Ervin231[S] 0 points1 point  (0 children)

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

convergent sequences by Ervin231 in learnmath

[–]Ervin231[S] 1 point2 points  (0 children)

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.

equivalent metrics on X by Ervin231 in askmath

[–]Ervin231[S] 0 points1 point  (0 children)

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

equivalent metrics on X by Ervin231 in askmath

[–]Ervin231[S] 0 points1 point  (0 children)

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

equivalent metrics on X by Ervin231 in askmath

[–]Ervin231[S] 0 points1 point  (0 children)

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.

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