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

  • #1

Homework Statement



I want to show that if [tex]C[/tex] and [tex]D[/tex] and closed sets, then [tex]C \bigcup D[/tex] 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 [tex]C[/tex] and [tex]D[/tex] are closed sets. Can I prove this statement by supposing that if [tex]c[/tex] is an accumulation point of [tex]C[/tex] and [tex]d[/tex] is an accumulation point of [tex]D[/tex], then both [tex]c[/tex] and [tex]d[/tex] will be accumulation points of [tex]C \bigcup D [/tex] ?
 

Answers and Replies

  • #2
CompuChip
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Yes. So you should start your proof with: let x be an accumulation point of [itex]C \cup D[/itex] and then show that [itex]x \in C \cup D[/itex]. This is pretty trivial.
 
  • #3
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
How does this proof look?

Proof: Suppose [tex]x_0[/tex] is an accumulation point of [tex]C \bigcup D[/tex]. Then either [tex]x_0[/tex] is an accumulation point of [tex]C[/tex] or [tex]x_0[/tex] is an accumulation point of [tex]D[/tex] or [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex].

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

Case II: Suppose [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex]. Then [tex]x_0[/tex] is an accumulation of [tex]C \bigcup D[/tex]. Therefore, [tex]C \bigcup D[/tex] is closed.
[tex]\Box[/tex]​
 
  • #5
HallsofIvy
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How does this proof look?

Proof: Suppose [tex]x_0[/tex] is an accumulation point of [tex]C \bigcup D[/tex]. Then either [tex]x_0[/tex] is an accumulation point of [tex]C[/tex] or [tex]x_0[/tex] is an accumulation point of [tex]D[/tex] or [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex].

Case I: Without loss of generality, suppose that [tex]x_0[/tex] is an accumulation point of [tex]C[/tex]. Since [tex]x_0[/tex] is an accumulation point of [tex]C[/tex], it follows that [tex]x_0[/tex] is an accumulation point of [tex]C \bigcup D[/tex]. Therefore, [tex]C \bigcup D[/tex] is closed.
NO, x0 is an accumulation point of [tex]C \bigcup D[/tex] by hypothesis. You wanted to prove it was IN [tex]C \bigcup D[/tex]

Case II: Suppose [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex]. Then [tex]x_0[/tex] is an accumulation of [tex]C \bigcup D[/tex]. Therefore, [tex]C \bigcup D[/tex] is closed.
[tex]\Box[/tex]​
Same point. In both cases, you need to show that x0 is IN [tex]C \bigcup D[/tex], 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
Ok, I see what I did wrong. Let me try this again. Thanks
 
  • #7
Does this look better?

Proof: Suppose [tex]x_0[/tex] is an accumulation point of [tex]C \bigcup D[/tex]. Then either [tex]x_0[/tex] is an accumulation point of [tex]C[/tex] or [tex]x_0[/tex] is an accumulation point of [tex]D[/tex] or [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex].

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

Case II: Suppose [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex]. Since both [tex]C[/tex] and [tex]D[/tex] are closed, it follows that [tex]x_0 \in C[/tex] and [tex]x_0 \in D[/tex]. Thus [tex]x_0 \in C \bigcup D[/tex]. Therefore, [tex]C \bigcup D[/tex] is closed.
[tex]\Box[/tex]​
 
  • #8
HallsofIvy
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Does this look better?

Proof: Suppose [tex]x_0[/tex] is an accumulation point of [tex]C \bigcup D[/tex]. Then either [tex]x_0[/tex] is an accumulation point of [tex]C[/tex] or [tex]x_0[/tex] is an accumulation point of [tex]D[/tex] or [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex].
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 [tex]C \bigcup D[/tex]. 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?

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

Case II: Suppose [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex]. Since both [tex]C[/tex] and [tex]D[/tex] are closed, it follows that [tex]x_0 \in C[/tex] and [tex]x_0 \in D[/tex]. Thus [tex]x_0 \in C \bigcup D[/tex]. Therefore, [tex]C \bigcup D[/tex] is closed.
[tex]\Box[/tex]​
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 [tex]C \bigcup D[/tex], does it follow that it is a "whatever" point either C or D no matter what "whatever" point means?
 
  • #9
HallsofIvy said:
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 [tex]C \bigcup D[/tex]. 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?
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.

If I say that x is a "whatever" point of [tex] C \bigcup D [/tex] , 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 [tex]x_0[/tex] is an accumulation point of both [tex]C[/tex] and [tex]D[/tex]
 
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