Prove: Limit Point of H ∪ K if p is Limit Point of H or K

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SUMMARY

The discussion centers on proving that if p is a limit point of the union of two sets H and K, then p must be a limit point of at least one of those sets. The proof begins by assuming that p is not a limit point of H, leading to the conclusion that there exists an open interval S around p that contains no points from H. Since p is a limit point of H ∪ K, it follows that S must contain points from K. The conversation emphasizes the importance of understanding open intervals in the context of real numbers, specifically within the framework of limit points.

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Jaquis2345
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Moved from technical forums, so no template
Summary: Definition: If M is a set and p is a point, then p is a limit point of M if every open interval containing p contains a point of M different from p.
Prove: that if H and K are sets and p is a limit point of H ∪ K,then p is a limit point of H or p is a limit point of K

In this proof I have assumed that p is not a limit point of H and went on to state that there exists some open interval S that contains p s.t. no element of H (other than possibly p itself) is in S. Since p is limit point of HUK a member of HUK must exist in (a,b) that member being K.
I am currently trying to prove that p is a limit point of K by letting some open interval V be any open interval containing p so S and V intersect but I can not seem to elaborate on what I have.
 
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Open interval means you are working in ##\mathbb{R}##?

Every open interval is also an interval around p when you intersect with the (a,b) that you found. This new interval must contain either an element of H or K. Which is is, and what does that say about your original interval?
 
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Office_Shredder said:
Open interval means you are working in ##\mathbb{R}##?

Every open interval is also an interval around p when you intersect with the (a,b) that you found. This new interval must contain either an element of H or K. Which is is, and what does that say about your original interval?
So no element of H can exist in the new interval other than possibly p. Thus, K exists in (a,b) and (c,d) where every point of K in that intersection is not equal to p. Right?
 
Jaquis2345 said:
In this proof I have assumed that p is not a limit point of H
Why not assume that ##p## is not a limit point of ##H## and not a limit point of ##K##? Then, try to show that it's not a limit point of ##H \cup K##?

That's called a proof by contraposition.
 
The student seems to be trying a proof by "partial converse." That is, he's assuming p is a limit point of H U K but not of H; he then intends to show it must be a limit point of K. But if some neighborhood of p contains (besides p) no point of H, but must contain a point of H U K, does that neighborhood not have to contain a point of K? This post is prompted by the desire to use the student's original idea, since he is already trying to use it.
 
Maybe this approach will be helpful: If p is a limit point of ##A\cup B ## the every open ##O_p## set/'hood containing p will intersect ##A \cup B##. The latter is the collection of points contained in ##A,B## or both.
Then if ##O_p## intersects neither of ##A,B##...
 

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