Using the Sequence Criterion for Closedness

[RVP91]

For a subset C is a subset of R^d
the following conditions are
equivalent:
(i) C is closed.
(ii) For every sequence (x_n) is a subset of C which converges in R^d,
the limit lim (as n goes to infinity) x_n = x must also lie in C.

Using the above theorem for the Sequence Criterion for Closedness. I need to show that the set C={(x,y) in R^2 | xy=a}, where a is in R, is a closed set of R^2.

Im finding it rather confusing.

So far I have said a point in the set must be of the form (a/y,a/x) but am finding it hard to write this as a sequence (z_n) which i can say is in C and then assume it is convergent in order to show its limit lies in C also and then conclude C is closed.

Any suggestions?

Thanks

 

[jgens]

Note that C is the graph of the function defined by f(x) = a/x. Then C is closed if and only if for every convergent sequence of points (x_n,f(x_n) we have lim_n→∞(x_n,f(x_n) in C. What does this force about f?

 

[RVP91]

Does it force f to be continuous? I'm not entirely sure how to explain why it needs to be continuous though, is it because it can only have a limit in C if it is continuous at all points in C?

 

[jgens]

Suppose X,Y are metric spaces. A function f:X → Y is continuous if and only if for every sequence of points x_n → x as n → ∞, we have f(x_n) → f(x) as n → ∞. This is sometimes taken as the definition of continuity. If you do not know this fact, try to prove it on your own, and if you have problems ask for help here :)

 

[micromass]

Post moved to homework forums.

 

[RVP91]

I'm quite confused as to where we at in showing C is closed.
From what I've followed so far we've concluded C is the graph of the function f(x)=a/x and so points in the set are (x,f(x)).
We then attempt to prove C is closed using the sequence criterion for closedness by considering the convergent sequence (x_n,f(x_n)). We have to show the limit of (x_n,f(x_n)) as n tends to infinity is also in C.
So can we let x = limit of (x_n,f(x_n)), then since x_n converges to x in R (using the assumption x_n converges we have f(x_n) converges to f(x) using the definition of continuity.

Is anything there correct?

Thanks again for the help

 

[jgens]

That's essentially the proof right there. The key is noticing that the graph of a function is closed if and only if the function is continuous. You just have to write out the relevant 'if and only if' statements to prove it. What you have above is the basic idea behind it, so you just need to flesh that out :)

 

[RVP91]

Could you possibly start to write the start of the proof? Or help me to understand where the relevant "if and only if" statements need to go please?
I just seem to have a few ideas scattered everyone and can't sort them into a proper proof apart from what I posted previously. Also I'm not even sure how you would reach the end conclusion that C is in fact closed.
The fact that x_n tends to x and f(x_n) tends to f(x), is that all we need to conclude it is closed or is there another step which makes it clear?

Thank you for helping

 

[jgens]

Let X,Y be metric spaces. Then f:X → Y is continuous if and only if lim_n→∞x_n = x implies lim_n→∞f(x_n) = f(x) if and only if lim_n→∞x_n = x implies lim_n→∞(x_n,f(x_n)) = (x,f(x)).

Now suppose (x_n,f(x_n)) is a convergent sequence in C. Then we know that lim_n→∞x_n exists. So what does the statement above tell you?

 

[RVP91]

Knowing limn→∞x_n exists tells us that limn→∞f(x_n) = f(x) which is true iff f is continuous?

 

[jgens]

Sure, but if you use the 'iff' statements I strung together above, you know even more. Look at the last clause in the 'iff' statement.

 

[jgens]

If you still don't get it:

Spoiler

Let X,Y be metric spaces. Then f:X → Y is continuous if and only if lim_n→∞x_n = x implies lim_n→∞f(x_n) = f(x) if and only if lim_n→∞x_n = x implies lim_n→∞(x_n,f(x_n)) = (x,f(x)).

Now suppose (x_n,f(x_n)) is a convergent sequence in C. Then we know that lim_n→∞x_n exists. If f is continuous, then lim_n→∞x_n = x implies lim_n→∞(x_n,f(x_n)) = (x,f(x)) by the statement above. Since (x,f(x)) in C, this proves every convergent sequence in C has limit in C. This proves that C is closed, completing the proof.

 

[RVP91]

Thank you.

Just another question. You said "if f continuous" in the proof. Is f in this case f(x) = a/x and so this is continuous?

 

[jgens]

In this case, f is the function defined by f(x) = a/x and f is continuous.

 

[RVP91]

So just to clarify is the following an adequate proof.

The set C is the graph of the continuous function y=f(x)=a/x. Thus a point in C has the form (x,f(x)).

Suppose (x_n,f(x_n)) is a convergent sequence in C. Then we know that limn->infinity x_n exists since x_n converges, let x = limn->infinty x_n. Then since f is continuous limn->infinity x_n = x implies limn->infinity f(x_n) = f(x) by the properties of continuous functions which implies limn->infinity (x_n, f(x_n)) = (x, f(x)).
Since (x,f(x)) is in C we have shown every convergent sequence in C has a limit in C(converges to a point in C). Hence C is closed.

Thanks again for the help!

 

[micromass]

Do notice that the above proof is insufficient if a=0. Because then your C is not the graph of a function anymore. So that requires another proof.

 

[RVP91]

Since C is given as C={(x,y) in R^2 | xy=a}. Does this mean I do indeed need to have a separate proof for when a = 0?