# Prove that if C and D are closed sets, then C U D is a closed set.

1. Oct 25, 2008

### opticaltempest

1. The problem statement, all variables and given/known data

I want to show that if $$C$$ and $$D$$ and closed sets, then $$C \bigcup D$$ is a closed set.

2. Relevant equations
A set is called closed iff the set contains all of its accumulation points.

3. The attempt at a solution

In order for me to prove this statement, I will be able to use the fact that $$C$$ and $$D$$ are closed sets. Can I prove this statement by supposing that if $$c$$ is an accumulation point of $$C$$ and $$d$$ is an accumulation point of $$D$$, then both $$c$$ and $$d$$ will be accumulation points of $$C \bigcup D$$ ?

2. Oct 25, 2008

### CompuChip

Yes. So you should start your proof with: let x be an accumulation point of $C \cup D$ and then show that $x \in C \cup D$. This is pretty trivial.

3. Oct 25, 2008

### opticaltempest

I wish this was trivial for me. :) Well, I had the correct intuition on how to prove the statement. Let me see if I can write the proof correctly. Thanks.

4. Oct 25, 2008

### opticaltempest

How does this proof look?

Proof: Suppose $$x_0$$ is an accumulation point of $$C \bigcup D$$. Then either $$x_0$$ is an accumulation point of $$C$$ or $$x_0$$ is an accumulation point of $$D$$ or $$x_0$$ is an accumulation point of both $$C$$ and $$D$$.

Case I: Without loss of generality, suppose that $$x_0$$ is an accumulation point of $$C$$. Since $$x_0$$ is an accumulation point of $$C$$, it follows that $$x_0$$ is an accumulation point of $$C \bigcup D$$. Therefore, $$C \bigcup D$$ is closed.

Case II: Suppose $$x_0$$ is an accumulation point of both $$C$$ and $$D$$. Then $$x_0$$ is an accumulation of $$C \bigcup D$$. Therefore, $$C \bigcup D$$ is closed.
$$\Box$$​

5. Oct 25, 2008

### HallsofIvy

NO, x0 is an accumulation point of $$C \bigcup D$$ by hypothesis. You wanted to prove it was IN $$C \bigcup D$$

Same point. In both cases, you need to show that x0 is IN $$C \bigcup D$$, not that it is an accumulation point- that was your hypothesis. Somewhere in there you will need to use the definition of "accumulation point".

6. Oct 25, 2008

### opticaltempest

Ok, I see what I did wrong. Let me try this again. Thanks

7. Oct 25, 2008

### opticaltempest

Does this look better?

Proof: Suppose $$x_0$$ is an accumulation point of $$C \bigcup D$$. Then either $$x_0$$ is an accumulation point of $$C$$ or $$x_0$$ is an accumulation point of $$D$$ or $$x_0$$ is an accumulation point of both $$C$$ and $$D$$.

Case I: Without loss of generality, suppose that $$x_0$$ is an accumulation point of $$C$$. Since $$C$$ is closed, it follows that $$x_0 \in C$$. Since $$x_0 \in C$$, it follows that $$x_0 \in C \bigcup D$$. Therefore, $$C \bigcup D$$ is closed.

Case II: Suppose $$x_0$$ is an accumulation point of both $$C$$ and $$D$$. Since both $$C$$ and $$D$$ are closed, it follows that $$x_0 \in C$$ and $$x_0 \in D$$. Thus $$x_0 \in C \bigcup D$$. Therefore, $$C \bigcup D$$ is closed.
$$\Box$$​

8. Oct 25, 2008

### HallsofIvy

Your teacher is likely to ask you to justify this. Suppose there were some sequence {xn}, with some points in C and some in D that converges to x0. Then x0 is an accumulation point of $$C \bigcup D$$. Does it follow that there must exist a sequence {cn} completely in C that converges to x0
or a sequence {dn} completely in D that converges to x0?

Once you have cleared up the point I mentioned, there is no need to do this second case. If x0 is an accumulation point of both C and D, then it is an accumulation point of C and Case I applies.

I notice that you still haven't used the definition of "accumulation point". If I say that x is a "whatever" point of $$C \bigcup D$$, does it follow that it is a "whatever" point either C or D no matter what "whatever" point means?

9. Oct 25, 2008

### opticaltempest

Well, I think this is true. Intuitively, I want to say that if there is a sequence {xn} with some points in both C and D that accumulate at x0, and we remove those points only in D, then we would still be able to find subsequences in C that converge to x0, and vice versa.

Regarding me not using the definition of accumulation point in this proof, I guess that is because not only am I weak with the concept of accumulation points, but I'm still learning how to write proofs. I wasn't sure if I even had to use the definition of an accumulation point. I'm still not sure if I have to use it.

I see that this is false. I was thinking of intersection when I wrote $$x_0$$ is an accumulation point of both $$C$$ and $$D$$

Last edited: Oct 25, 2008