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

In summary, the conversation discusses the proof of a statement that states if C and D are closed sets, then C \bigcup D is also a closed set. The conversation includes the attempt at a solution and a suggestion to start the proof with the assumption that x is an accumulation point of C \bigcup D. The conversation also includes a discussion on how to justify this assumption and a revised proof using the definition of accumulation points.
  • #1
opticaltempest
135
0

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] ?
 
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  • #2
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
opticaltempest said:
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
opticaltempest said:
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]
 
Last edited:

1. What does it mean for a set to be "closed"?

A set is considered "closed" if it contains all of its limit points. This means that if a sequence of points within the set converges, the limit of that sequence must also be within the set.

2. Why is it important for a set to be closed?

Having closed sets is important in mathematical analysis because it allows for more precise calculations and proofs. Closed sets have well-defined boundaries and properties, making them easier to work with.

3. How do you prove that C U D is a closed set if C and D are closed sets?

To prove that C U D is a closed set, we must show that it contains all of its limit points. This can be done by assuming that there is a limit point of C U D that is not in the set, and then using the definitions of limit points and closed sets to arrive at a contradiction.

4. Can you provide an example to illustrate this theorem?

Yes, for example, let C = [0,1] and D = [2,3]. Both C and D are closed sets, as they contain all of their limit points. Now, C U D = [0,1] U [2,3] = [0,3], which is also a closed set because it contains all of its limit points (0 and 3).

5. Is this theorem applicable to all types of sets?

No, this theorem is specifically applicable to closed sets. It does not hold true for all types of sets, such as open or compact sets. However, it can be extended to other types of sets by modifying the definitions and proof techniques accordingly.

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