Does {1/n} n=1 to infinity converge? Why or why not? in a topological space

[math25]

hi,
can someone please help me with this problem.

Let T be the collection of all U subset R such that U is open using the usual
metric on R.Then (R; T ) is a topological space. The topology T could also be described as
all subsets U of R such that using the usual metric on R, R \ U is closed and
bounded.

Does {1/n} n=1 to infinity converge? Why or why not?

I think it does converge...it converges to 1 for example...am I right?

Does {n} n=1 to infinity converge? Why or why not?

I don't think that this sequence converges in a topological space?

thanks

 

[jgens]

Does {1/n} n=1 to infinity converge? Why or why not?

I think it does converge...it converges to 1 for example...am I right?

Does there exist an $N \in \mathbb{N}$ such that the open neighborhood $(\frac{1}{2},\frac{3}{2})$ of $1$ contains $\frac{1}{n}$ for all $N<n$?

Does {n} n=1 to infinity converge? Why or why not?

I don't think that this sequence converges in a topological space?

You should say "I don't think that this sequence converges in the usual topology on $\mathbb{R}$" to be more accurate. But you got this one right. Now can you prove it?

 

[math25]

thank you so much...

It seems like I wrote everything wrong, for the second sequence that's what I meant to say, and I've already proved it.

for the first sequence, it converges to 0 in usual topology and this is what I have so far...

For every open U s.t. 0 is in U, there exist N such that for all n > N, 1/n is in U

Assume 0 is in U where U is open in usual topology then there is a s> 0 s.t. (1-s, 1+s) is subset of U
So,

1-s < 1/n
n < 1/( 1-s)

 

[jgens]

For every open U s.t. 0 is in U, there exist N such that for all n > N, 1/n is in U

Assume 0 is in U where U is open in usual topology then there is a s> 0 s.t. (1-s, 1+s) is subset of U
So,

1-s < 1/n
n < 1/( 1-s)

Given that you wrote the first part wrong, I am going to ask you to rewrite this argument carefully before I give you more help.

 

[HallsofIvy]

hi,
can someone please help me with this problem.

Let T be the collection of all U subset R such that U is open using the usual
metric on R.Then (R; T ) is a topological space. The topology T could also be described as
all subsets U of R such that using the usual metric on R, R \ U is closed and
bounded.

Does {1/n} n=1 to infinity converge? Why or why not?

I think it does converge...it converges to 1 for example...am I right?

Did you think to write out at least part of sequence? Taking n= 1, 2, 3, 4, 5, gives
1, 1/2, 1/3, 1/4, 1/5. You think that is getting closer to 1?

Does {n} n=1 to infinity converge? Why or why not?

I don't think that this sequence converges in a topological space?

thanks

 

[math25]

It was mistake, which I corrected later...it converges to 0 not 1.

 

[jgens]

It was mistake, which I corrected later...it converges to 0 not 1.

Right. But you still have not provided a correct proof that the sequence converges to 0.

 

[math25]

agree, I"ll be working on it today...