Prove that A is closed

• Nov 13th 2011, 11:20 AM
wopashui
Prove that A is closed
Suppose that $x_n-->x$ $\in$ $M$ on in metric d, and let A={x}U { $x_n:n>=1$}. Prove that A is closed.

intuitively, x is a limit point of $x_n$, and it's inside A, so A must be closed, but how do I write out this proof?
• Nov 13th 2011, 11:32 AM
Drexel28
Re: Prove that A is closed
Quote:

Originally Posted by wopashui
Suppose that $x_n-->x$ $\in$ $M$ on in metric d, and let A={x}U { $x_n:n>=1$}. Prove that A is closed.

intuitively, x is a limit point of $x_n$, and it's inside A, so A must be closed, but how do I write out this proof?

It's actually probably easiest to prove that $A$ is compact, and to use the fact that in metric spaces (Hausdorff spaces) compactness implies closedness. To prove that $A$ is compact you must merely note that given an open cover $\Omega$ for $A$ you can find some guy that contains $x$, but since $x_n\to x$ this guy will contain all but finitely many elements of $A$ and you can cover the finitely many rest with whatever--this gives you a finite subcover.
• Nov 13th 2011, 12:11 PM
Tinyboss
Re: Prove that A is closed
It's immediately true if you know that a convergent sequence only has one limit point, which isn't hard to show. Choose y not equal to x, and not in {x_n}. By Hausdorff-ness of the metric space, there are disjoint neighborhoods of y and x. By convergence, all but finitely many elements of the sequence lie in the chosen neighborhood of x, and so y can't be a limit point.
• Nov 14th 2011, 04:03 AM
wopashui
Re: Prove that A is closed
Quote:

Originally Posted by Tinyboss
It's immediately true if you know that a convergent sequence only has one limit point, which isn't hard to show. Choose y not equal to x, and not in {x_n}. By Hausdorff-ness of the metric space, there are disjoint neighborhoods of y and x. By convergence, all but finitely many elements of the sequence lie in the chosen neighborhood of x, and so y can't be a limit point.

what is Hausdorff-ness? and we havent not define compactness yet
• Nov 14th 2011, 04:11 AM
Plato
Re: Prove that A is closed
Quote:

Originally Posted by wopashui
what is Hausdorff-ness? and we havent not define compactness yet

Well, if your notes do not cover Hausdorff spaces or compactness then in the notes is it proved that limits of sequences are unique?
If not, what have you shown with regard to sequences and limit points?
• Nov 14th 2011, 04:36 AM
Tinyboss
Re: Prove that A is closed
Sorry, Hausdorff just means that for any two distinct points we can find an open set containing each one, and so that the two open sets don't intersect one another. You usually don't encounter that definition until you consider more general topological spaces, because metric spaces are automatically Hausdorff (just take epsilon-balls of radius half the distance between the points).
• Nov 14th 2011, 04:41 AM
Opalg
Re: Prove that A is closed
Quote:

Originally Posted by wopashui
Quote:

Originally Posted by Tinyboss
It's immediately true if you know that a convergent sequence only has one limit point, which isn't hard to show. Choose y not equal to x, and not in {x_n}. By Hausdorff-ness of the metric space, there are disjoint neighborhoods of y and x. By convergence, all but finitely many elements of the sequence lie in the chosen neighborhood of x, and so y can't be a limit point.

what is Hausdorff-ness? and we havent not define compactness yet

You can take Tinyboss's argument and adapt it to metric spaces.

Choose y not equal to x, and not in {x_n}. Then d(x,y)>0. The open ball of radius d(x,y)/2 centred at x contains all but finitely many elements of the sequence. So the open ball of radius d(x,y)/2 centred at y (which is disjoint from the one centred at x) contains only finitely many members of the sequence. By shrinking its radius to exclude those points, you get an open ball centred at y which does not contain any points of A. That shows that A is closed.
• Nov 14th 2011, 04:35 PM
wopashui
Re: Prove that A is closed
Quote:

Originally Posted by Opalg
You can take Tinyboss's argument and adapt it to metric spaces.

Choose y not equal to x, and not in {x_n}. Then d(x,y)>0. The open ball of radius d(x,y)/2 centred at x contains all but finitely many elements of the sequence. So the open ball of radius d(x,y)/2 centred at y (which is disjoint from the one centred at x) contains only finitely many members of the sequence. By shrinking its radius to exclude those points, you get an open ball centred at y which does not contain any points of A. That shows that A is closed.

for "By shrinking its radius to exclude those points, you get an open ball centred at y which does not contain any points of A. That shows that A is closed" this part, can you explain expiciltly how do I do that, I think that's the tricky part, how to choose that radius, let's say we have r=d(x,y), for the 2 open ball we have radius of r/2, but hot can you shrink the radius to be small enough to do this?
• Nov 14th 2011, 04:45 PM
Tinyboss
Re: Prove that A is closed
Because only finitely many sequence elements are outside the ball around x, you can pick out a closest sequence element to y. Just make your radius smaller than that distance.