Thread: compact set and convergent sequence

1. 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.

2. 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.

3. This is of course true in any topological space.

4. 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 $\displaystyle p_n$.