# Basic topology

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• Mar 4th 2010, 12:57 PM
mohammadfawaz
Basic topology
Hello,

I have the following problem:
Prove that the set of limit points E' of a subset E is a closed set.

Help please,

Thank you,
• Mar 4th 2010, 01:02 PM
mabruka
What definition of closed are you using?

I think the complement of an open set right?
• Mar 4th 2010, 01:11 PM
Plato
Quote:

Originally Posted by mohammadfawaz
Prove that the set of limit points E' of a subset E is a closed set.

This may not be true in a general space.
It depends upon the nature of the space.
What sort of topological space is this set in?
• Mar 4th 2010, 03:29 PM
Drexel28
Quote:

Originally Posted by mohammadfawaz
Hello,

I have the following problem:
Prove that the set of limit points E' of a subset E is a closed set.

Help please,

Thank you,

This is true in $T_1$ spaces. Maybe that is what the OP meant?
• Mar 4th 2010, 04:40 PM
southprkfan1
Quote:

Originally Posted by mohammadfawaz
Hello,

I have the following problem:
Prove that the set of limit points E' of a subset E is a closed set.

Help please,

Thank you,

Let (xn) be a convergent sequence in E', (xn) --> x. We want to show x is in E'. I.e, we want to show x is a limit point of E.

We know x is a limit point of E if given any e>0 there is a point in E that is in the ball of radius e around x. That is, there exists a p in E such that:
d(x,p) < e

Heres the proof:

Since (xn) --> x, then there is a k where d(x,xk) < e/2 *

and since xk in in E', it is a limit point of E, so there is a point p in E such that: d(xk,p) < e/2 **

I claim: d(x, p) < e

Pf.

d(x,p) <= d(x,xk) + d(xk, p) < e/2 + e/2 (by * and **)
so d(x, p) < e

Edit: Put Metric d instead of absolute value, if that's not what you mean, just replace every d(x,y) with lx - yl
• Mar 4th 2010, 04:41 PM
Drexel28
Quote:

Originally Posted by southprkfan1
Let (xn) be a convergent sequence in E', (xn) --> x. We want to show x is in E'. I.e, we want to show x is a limit point of E.

We know x is a limit point of E if given any e>0 there is a point in E that is in the ball of radius e around x. That is, there exists a p in E such that:
lx - pl < e

Heres the proof:

Since (xn) --> x, then there is a k where l x-xk l < e/2 *

and since xk in in E', it is a limit point of E, so there is a point p in E such that: l xk - p l < e/2 **

I claim: lx - pl < e

Pf.

lx - pl <= lx-xkl +lxk - pl < e/2 + e/2 (by * and **)
so lx- pl < e

....in a metric space.