# Math Help - Metric Space

1. ## Metric Space

Need help proving the following statements/giving counter-examples;

(X,d) is a metric space

1: for all a
єX, U n=1 --> ∞, B(1/n, a) is an open set
2:
for all a
єX, U n=1 --> ∞, B(1/n, a) is not an open set
3: if xn --> x, then any subsequence of xn converges to x
4: if any subsequence of xn converges to x, then xn converges to x
5: if U is open then U=Int(Uconjugate)

So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
ε greater than the radius of 1/n so it is open...is this right?

2. Originally Posted by math_help
Need help proving the following statements/giving counter-examples;

(X,d) is a metric space

1: for all a
єX, U n=1 --> ∞, B(1/n, a) is an open set
2:
for all a
єX, U n=1 --> ∞, B(1/n, a) is not an open set
3: if xn --> x, then any subsequence of xn converges to x
4: if any subsequence of xn converges to x, then xn converges to x
5: if U is open then U=Int(Uconjugate)

So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
ε greater than the radius of 1/n so it is open...is this right?
I don't understand what you mean by "you can always choose $\epsilon$ greater than 1/n so it is open". WHAT is open? For every n, of course B(1/n,a) is an open set because all such neighborhoods are open. But to prove that you would need to assert that you can always choose [tex]\epsilon[/itex] less than 1/n, not greater than. In any case, (1) and (2) if the question is whether they are true for all metrics, then that statement is false. $\lim_{x\rightarrow a}B(1/n, a)= {a}$. In the "discrete" metric (d(a,b)= 1 if a is not equal to b, 0 otherwise) all singleton sets are both open and closed so (2) is not true but if in, say, R, d(a,b)= |a- b|, then all singleton sets are closed but not open so (2) in that case is true.

As far as (4) is concerned the English word "any" is ambiguous. I could see interpreting this as "If every subsequence... " in which case it is true. I could also see interpreting it as "if some subsequenc ..." in which case it is false.

For (5), I don't know what "Uconjugate" means. Did you mean U complement? In that case it is certainly not true! Or did you mean "Uclosure"? In that case it is true.

3. Sorry, never stated some things clearly.

It simply says "everywhere (X,d) is a metric space" so i'd guess the question is implying it can be any metric space.

For 1 and 2 I mean Union, but it looks like you knew that one. 5 isn't union and yes U compliment sorry, stupid mistake there.

Yeah 4 is worded badly, my interpretation is that the question is "if there exists a of subsequence xn that converges to x, then xn also converges to x.

So 3 is true, how can I go about proving this? and for number 4 i'm confused to the concept of a sequence in a metric space to be honest, naturally I want to say sin(k*pi/2) is subsequence of sin which converges for 1, where K is an integer, however sin doesn't converge, but can't get my head around it. Thanks in advance for any tips, and the original ones!

4. For #3, do you fully understand the definition of subsequence?
There must be a increasing sequence of positive integers, $\alpha (n)$, such that $\left\{ {x_{\alpha (n)} } \right\} \subseteq \left\{ {x_n } \right\}$.
If $p$ is the limit point of $x_n$ then for any open set $O$ if $p \in O$ then $O$ contains almost all the terms of $x_n$.
So is that not also true of any subsequence?

5. Originally Posted by math_help
Need help proving the following statements/giving counter-examples;

(X,d) is a metric space

1: for all a єX, U n=1 --> ∞, B(1/n, a) is an open set
2: for all a
єX, U n=1 --> ∞, B(1/n, a) is not an open set
3: if xn --> x, then any subsequence of xn converges to x
4: if any subsequence of xn converges to x, then xn converges to x
5: if U is open then U=Int(Uconjugate)

So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
ε greater than the radius of 1/n so it is open...is this right?

Statement 4 is false ,as the following counter example shows:

Take the sequence : 1,-1,1,-1,1,-1............................the sequence does not converge, it only has 1,-1 as accumulation points.

But every subsequence of the sequence has the limit 1 or -1

6. Originally Posted by benes
Statement 4 is false ,as the following counter example shows: Take the sequence : 1,-1,1,-1,1,-1............................the sequence does not converge, it only has 1,-1 as accumulation points.
But every subsequence of the sequence has the limit 1 or -1
Sorry, but that is not a counterexample to the statement.
Did you carefully read the statement if the statement?
“any subsequence of $x_n$ converges to $\color{blue}x$.
$\color{blue}x$ is one number not two as in your so called example.

I also disagree with an earlier post. Surely $x_n$ is a subsequence of itself.
So part 4 is true!

7. Originally Posted by Plato
Sorry, but that is not a counterexample to the statement.
Did you carefully read the statement if the statement?
“any subsequence of $x_n$ converges to $\color{blue}x$.
$\color{blue}x$ is one number not two as in your so called example.

I also disagree with an earlier post. Surely $x_n$ is a subsequence of itself.
So part 4 is true!
YES you right i misread the question,thank you.

But which earlier post??