@JasperLoy The section has only one theorem $(H,d)$ is a metric space, where $$d(x,y)=\left[\sum_{i=1}^{\infty}(x_i-y_i)^2 \right]^{1/2}$$

and $H$ is the space of all sequences of reall numbers such that $\sum_{i=1}^\infty u_i^2$ converges.

@JasperLoy Well, I think that is related to topology then.

@JasperLoy Remember what I was doing yesterday?

@PeterTamaroff If you are enjoying working through Mendelson, that is probably more important than any advice given by the rest of us - go for it!

@OldJohn =D

@JasperLoy Dude, you there?

@JasperLoy I writing the proof that if $Y$ is a subspace of $X$ then $N'\in Y$ is a nieghoborhood of $a\in Y$ $\iff$ there exists a neighborhood $N$ of $a$ in $X$ such that $N\cap Y=N'$.

Now, I think this is one part:

@JasperLoy Yes.

Suppose $N'\subset Y$ is a nbhd of $a\in Y$. Then $B_Y(a;\delta)\subset N'$ where $B_Y$ is an open ball in $Y$. Consider the same open ball in $X$, this is $B_X(a;\delta)$. Then $B_X$ is a neighborhood of $a$.

However, what I have is $B_X\cap Y=B_Y$ here...

I need to show there is an $N$ such that $N\cap Y=N'$.

Balls form a basis for the metric topology

@HenryT.Horton Yes.

And you just showed it is true for balls. So...

@HenryT.Horton RIght, But then I have to change the wording of that.

Well whatever "that" is, go ahead and change it

@HenryT.Horton Well, I should star with "Since open balls about $a$ form a basis for the system of nbhds at $a$, it suffices to prove the result for open balls."

Sure

@JasperLoy An open ball in $Y$ is $B_Y(a;\delta)=\{x\in Y:d(x,a)<\delta\}$

The "same" ball in $X$ is $B_X(a;\delta)=\{x\in X:d(x,a)<\delta\}$

Obviously this is a superset of the former ball.

@JasperLoy OK. So it is clear $B_X\cap Y=B_Y$

@JasperLoy What was the motivation of this exercise?

How about a decent kick in the groin?

Benjamin Lim mentined something about this being the definition of topology, or something of the sort.

This exercise is about the subspace topology

@JasperLoy 8-)?

If $X$ is a topological space and $Y \subseteq X$, then the subspace topology on $Y$ has open sets $U \cap Y$, where $U$ is an open set in $X$

So you really just showed that given a metric space $(X,d)$ and $Y \subset X$, the metric topology on $(Y, d|_{Y \times Y})$ is just the subspace topology on $Y$ induced from the metric topology on $X$

@HenryT.Horton Cool, thanks.

@HenryT.Horton I still wondering how to prove the result with nbhds

I proved it one way.

That is, if $N$ is a nbhd of $a$ in $X$, then $N\cap Y=N'$ is a nbhd of $a$ in $Y$.

That is pretty striaght forward.

NOw I want to prove that if $N'$ is a nbhd of $a$ in $Y$ then there must exists a nbhd $N$ of $a$ in $X$ such that $N'=N\cap Y$. Mabye I can argue by contrapositive.

@JonasTeuwen Hi Jonas - sounds pretty indecent to me :P

You showed it that way....

With balls

@OldJohn :P.

$B_Y(a;\delta) = Y \cap B_X(a,\delta)$

Ah - the conversation is all about balls now :)

@HenryT.Horton But can it be proven in the generality of nbhds?

Just play with the balls, Peter

Yeah, I like to play with balls. Double and bound them.

@HenryT.Horton HAHAHHAHAHAHAHHA

Don't go looking around the neighborhood for more

$\frac{8}{3}\pi r^3$ !!

@HenryT.Horton One thing.

Ahh, the ballume of a pair of balls

When I say that open balls form a basis for the system of nbhds at $a$ I mean any nbhd of $a$ contains a point of an open ball about $a$.

Assuming they have equal radius

Peeter, do you know that any neighborhood is a union of ballz?

Balls usually have a slightly different radius because that way they don't get squished when you cross your legs.

@HenryT.Horton I know any open set is an union of balls.

@JonasTeuwen Too much information

@OldJohn There is no such thing 8-).

@JonasTeuwen :)

A neighborhood is in particular an open set

@HenryT.Horton I have defined a nbhd of $a$ as a set containing an open ball about $a$.

@HenryT.Horton How can you prove that? Is is true for metric spaces?

My definition of a neighborhood of $a$ is an open set containing $a$

Because I'm being asked to prove analogous results for open sets, closed sets and neighborhoods.

@JasperLoy Well, open balls form a basis for the system of nbhds at $a$ because given any nbhd of $a$, it contains a point of an open ball about $a$.

(other than $a$)

Harmonic analysts just play with the balls, bro.

I made that ball joke before :-(. I need more. Need to regenerate my brain.

What did I do this time, Sassper?

@JasperLoy Good fun eh...?

@JasperLoy How would you explain why it suffices to? I forgot why.

@JasperLoy Well, I proved it directly in one way.

Then I ued the balls argument the other. I can't see hwo to prove the other way without my precious balls.

@JasperLoy What is your opinion on this exercise: "Consider the subspace $(\BBb Q,d\mid \BBb Q\times \BBb Q)$ of $(\Bbb R,d)$. Let $\{a_n\}$ be a sequence of rational numbers such that $\lim \; a_n=\sqrt{2}$. Prove that given $\epsilon 0$, there is a positive integer $N$ such that for $n,m>N$, $|a_n-a_m|<\epsilon$.

What do you mean opinion

The exercise shows that $(\Bbb Q, d_{\Bbb Q \times \Bbb Q})$ is not a complete metric space

@HenryT.Horton He likes to critique books.

He also likes Justin Bieber, apparently.

@HenryT.Horton Doesn't it ask to show the sequence is Cauchy?

@HenryT.Horton LOL

Yes

And a complete metric space is a metric space where every Cauchy sequence converges to a point in that space

But $\sqrt{2} \notin \Bbb Q$

@HenryT.Horton Oh, OK.

And also, every convergent sequence is Cauchy.

Yes

The inequiality should follow by the triangle inequality, right?

Yes

Choosing $|\sqrt 2 -a_n|< \epsilon /2$, $|\sqrt 2 -a_m|< \epsilon /2$

or something of the sort.

@JasperLoy But we haven't even mentinoed Cauchy sequences.

@JasperLoy BYes

@HenryT.Horton He might be just inactive.

He is inert.