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

[opticaltempest]

Homework Statement

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

Homework Equations

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

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 ?

 

[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.

 

[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.

 

[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​

 

[HallsofIvy]

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.

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

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​

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".

 

[opticaltempest]

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

 

[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​

 

[HallsofIvy]

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.

Your teacher is likely to ask you to justify this. Suppose there were some sequence {x_n}, 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 {c_n} completely in C that converges to x₀
or a sequence {d_n} completely in D that converges to x₀?

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​

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?

 

[opticaltempest]

Your teacher is likely to ask you to justify this. Suppose there were some sequence {x_n}, 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 {c_n} completely in C that converges to x₀
or a sequence {d_n} completely in D that converges to x₀?

Well, I think this is true. Intuitively, I want to say that if there is a sequence {x_n} 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.

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?

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