# compact set and convergent sequence

• Feb 28th 2010, 01:00 PM
inthequestofproofs
compact set and convergent sequence
Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."

I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.

Proof:
Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.

I don't think my proof is clear at all. Please provide some hints.
• Feb 28th 2010, 01:39 PM
southprkfan1
Quote:

Originally Posted by inthequestofproofs
Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."

I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.

Proof:
Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.

I don't think my proof is clear at all. Please provide some hints.

Try this:

Let O be an open cover of B. O = {O_j, j=1,2,3...} And Suppose p is on O_k.

then there exists an e>0 such that the open ball of radius e around p is in O_k (by defintion of an open set)

but since (pn) --> p, there exists a number N such that for all n > N, we have l pn - p l < e

that is, for all n>N, pn is in the ball of radius e around p, and thus O_k is a cover for all pn, n>N.

Now we only have to worry about the points pn, n = 1,2,3...N. But, this is a finite number of points which is cleary covered by a finite number of open sets...and the rest is easy.
• Feb 28th 2010, 02:01 PM
Drexel28
This is of course true in any topological space.
• Feb 28th 2010, 02:26 PM
Plato
Quote:

Originally Posted by Drexel28
This is of course true in any topological space.

Yes it is true in general.
If $\displaystyle (p_n)\to p$ then for any open set $\displaystyle \mathcal{O}$ containing $\displaystyle p$ also contains almost all (all but a finite collection) points of $\displaystyle p_n$.