# Homework Help: Show that lim S is a closed set

1. Sep 14, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Suppose that pn → p and each pn lies in lim S. We claim that p ∈ lim S. Since
pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
Thus there exists qn = pn,k(n)∈ S such that
d(pn, qn) <1/n.
Then, as n→∞ we have
d(p, qn) ≤ d(p, pn) + d(pn, qn) → 0
which implies that qn → p, so p ∈ lim S, which completes the proof that lim S is a
closed set.

2. Relevant equations

3. The attempt at a solution
The only part of the proof I don't understand is the following:

Since
pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
Thus there exists qn = pn,k(n)∈ S such that
d(pn, qn) <1/n.

Where does the 1/n come from? Why can we be sure that such a qn exists?

2. Sep 14, 2017

### Dick

You are given that $p$ is a limit of points of $lim S$. You want to show that $p$ itself is an element of $lim S$. To do this you have show that there is a sequence of point in $S$ whose limit is $p$. The argument shows that for every $n$ there is a point of $S$ whose distance from $p$ is less than $1/n$. Which goes to zero as $n \rightarrow \infty$. Hence $p$ is in $lim S$. Clearer?

3. Sep 14, 2017

### FactChecker

Anything that goes to 0 will do the trick. You can use 1/n and n→∞ or you can use ε>0 and ε→0. They will both make the distance go to 0.
Because pn is a limit of points in S, there is a qn at least that close to pn.

4. Sep 16, 2017

### Mr Davis 97

I guess my main confusion lies in the following: Thus there exists qn = pn,k(n)∈ S such that d(pn, qn) <1/n.

Does this somehow come from the definition of convergence? Also, I am confused by the notation. The proof says that there is a sequence that converges to pn. So are we saying that n is some fixed number in this instance? Because obviously a sequence can't converge to another sequence.

5. Sep 17, 2017

### FactChecker

Yes. It comes from the sentence before: pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
For that statement, n is fixed. But the logic will work for any n, which is important since n →∞.
I agree. and it doesn't even have to converge to the same limit as the other sequence. It is converging to the limit of the points, pn that the sequences (pn,k)k∈N converge to.

6. Sep 17, 2017

### FactChecker

Start with the blue sequence of points, pn, converging to the red point, A. Each blue point is in S, so there is a sequence of green points converging to each one. Select the sequence of circled points converging to A. That circled sequence proves that A is also in S.