Compactness; show inf is achieved

[kingwinner]

Homework Statement

Homework Equations

N/A

The Attempt at a Solution

Since we're in Rⁿ, from the given information, we can say that X is compact.
But I don't really understand the hint in part a(actually I'm really confused). HOW can we find such sequences x_n and y_n? And how can we find a subsequence x_{n_k} that converges?

Any help is appreciated!


[note: also under discussion in math help forum]

 

[tt2348]

Ok I think the best way to approach this is by starting component wise.
By thinking of xn as the cross product of a whole lot of closed bounded intevals on R, you can consider each individual interval seperately. Call the interval of interest I_{k}

pick y_{n},x_{n} to converge to a point (a,b) on it's interval product.
now consider a subsequence of x_{n} such that for every y_{n}
d(x_{n_{k}},y_{n}) is a minimum

 

[tt2348]

then by picking y_(n_(k_m))) in such a way that d(y(j),x(nk)) is a min, for all n_k.
it'll be an inf for d(xn,yn) since x was picked so that it was the shortest distance for every point yn, and y was picked so that it was the shortest distance for every point of the newely refurbished series xnk. and since it was picked to converge, the subsequence converges to the same point.

 

[tt2348]

the hint was worthless though I won't lie.

 

[kingwinner]

I don't quite understand your point, so maybe I should go back to the hint...

I am not sure how to begin because I don't know how to FIND the x_n and y_n in the hint in the first place? Do we need to produce an actual concrete example for which d(x_n,y_n)->d(X,Y)? Or can we just say such sequences exist?

Also, since X is compact, and by definition we know there EXISTS a convergence subsequence x_nk whose limit lies in X. But then the hint tells me to FIND such a subsequence, why??

Can someone kindly explain this?
Thanks a million!

 

[tt2348]

You pick a pair of sequences converging to a point. You pick a subsequence so that d(y1,xn1) is a min, d(y2,xn2) is min.. Etc
and from their, you can make it even smaller by consider the values of a sequence ynkm where
d(ynk1,xn1) is min, d(ynk2,xn2) is a min... These sequence are guarenteed by compactness .

 

[tt2348]

You basically say that their are 2 sequences that satisfy it, and you show why it's true

 

[kingwinner]

Does anyone have any idea about part b??

 

[some_dude]

Does anyone have any idea about part b??

Any countable set (in \mathbb{R}^n) is closed. So just choose two sequences Which converge to the same point from different sides.

E.g., let
X = \{ \frac{1}{n} : n \in \mathbb{N} \}
Y = \{ -\frac{1}{n} : n \in \mathbb{N} \}

 

[kingwinner]

Any countable set (in \mathbb{R}^n) is closed. So just choose two sequences Which converge to the same point from different sides.

E.g., let
X = \{ \frac{1}{n} : n \in \mathbb{N} \}
Y = \{ -\frac{1}{n} : n \in \mathbb{N} \}

But I thought for part b we're supposed to find an example where X and Y are unbounded, and the infinmum is never achieved, no?

In your example, both X and Y are bounded...Also, I don't get why the infinmum is NOT achieved in your example. How can we prove it?

Thank you!

 

[some_dude]

But I thought for part b we're supposed to find an example where X and Y are unbounded, and the infinmum is never achieved, no?

In your example, both X and Y are bounded...Also, I don't get why the infinmum is NOT achieved in your example. How can we prove it?

Thank you!

Oh sorry! Let me correct myself!

In the metric space \mathbb{R}^2, let

<br /> X = \{ (\ n,\ 1/n) : n \in \mathbb{N} \}<br />
<br /> Y = \{ (\ n,-1/n) : n \in \mathbb{N} \}<br />

i.e., the graphs of what I told you before

 

[kingwinner]

Oh sorry! Let me correct myself!

In the metric space \mathbb{R}^2, let

<br /> X = \{ (\ 1/n,\ 0) : n \in \mathbb{N} \}<br />
<br /> Y = \{ (-1/n,\ 0) : n \in \mathbb{N} \}<br />

i.e., the graphs of what I told you before

Hi, but still your X and Y are bounded...And I don't understand why the infinmum is NOT achieved in your example. How to justify it?

thanks.

 

[some_dude]

Hi, but still your X and Y are bounded...And I don't understand why the infinmum is NOT achieved in your example. How to justify it?

thanks.

I just edited it look again... I The infimum is 0. But there exist no points in X, Y that satisfy this.

(This is just the graph of 1/n and -1/n going from 0 to infinity - both converge to 0, but for never actually take that value.)

 

[kingwinner]

I just edited it look again... I The infimum is 0. But there exist no points in X, Y that satisfy this.

(This is just the graph of 1/n and -1/n going from 0 to infinity - both converge to 0, but for never actually take that value.)

But how can we PROVE that d(X, Y) = inf{d(x, y) : x E X; y E Y} = 0 ?
I have no idea how to prove this...:(

 

[some_dude]

But how can we PROVE that d(X, Y) = inf{d(x, y) : x E X; y E Y} = 0 ?
I have no idea how to prove this...:(

To show d(X, Y) = 0:
1. note that d(X,Y) is bounded below by zero
2. show for an arbitrary epsilon > 0, there exists elements in X and Y that are closer than epsilon from each other
It follows from that.

To show show that for all x in X, y in Y that d(x,y) > 0, note the intersection of the closures of X and Y is empty (both are closed themselves and clearly disjoint). Apply the definition of the limit points of a set and it follows.

I need to go out for an hr, if you have another question, i'll check later. See ya.

 

[kingwinner]

To show d(X, Y) = 0:
1. note that d(X,Y) is bounded below by zero
2. show for an arbitrary epsilon > 0, there exists elements in X and Y that are closer than epsilon from each other
It follows from that.

OK, so 0 is a lower bound.
We have to prove that 0 is the GREATEST lower bound. But it seems like you're talking about convergence, and in R² (unlike in R), we don't have the monotone convergence theorem which links the greatest lower bound with convergence...
So how can we prove that 0 is the GREATEST lower bound?

To show show that for all x in X, y in Y that d(x,y) > 0, note the intersection of the closures of X and Y is empty (both are closed themselves and clearly disjoint). Apply the definition of the limit points of a set and it follows.

To show that d(x,y) > 0, can we instead say that since d(x,y) is a metric, d(x,y)=0 <=> x=y
And in your example, x E X and y E Y are never equal, so d(x,y)>0??


Thanks!

 

[some_dude]

OK, so 0 is a lower bound.
We have to prove that 0 is the GREATEST lower bound. But it seems like you're talking about convergence, and in R² (unlike in R), we don't have the monotone convergence theorem which links the greatest lower bound with convergence...
So how can we prove that 0 is the GREATEST lower bound?

The set A = \{ D(x,y) : x \in X, y \in Y \} is a subset of \mathbb{R}. If for an arbitrary \epsilon &gt; 0 you can show (and you can, easily) that it is not a lower bound for A it follows that inf A = 0 (as we know 0 is a lower bound).

To show that d(x,y) > 0, can we instead say that since d(x,y) is a metric, d(x,y)=0 <=> x=y
And in your example, x E X and y E Y are never equal, so d(x,y)>0??
Thanks!

Yes, I think that's fine. Maybe when you write it up add a one line explanation as to why X and Y are disjoint (don't know how picky your marker is).

 

[kingwinner]

The set A = \{ D(x,y) : x \in X, y \in Y \} is a subset of \mathbb{R}. If for an arbitrary \epsilon &gt; 0 you can show (and you can, easily) that it is not a lower bound for A it follows that inf A = 0 (as we know 0 is a lower bound).

I know it's easy for you, but not for me:(
If you don't mind, can you explain how to prove that ANYTHING >0 is not a lower bound.

d(x,y) = ||x-y|| = (n-n)² + [1/n-(-1/n)]² = 4/n² < ε

Is that how you prove that the greatest lower bound is 0?

Thanks!