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

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The discussion focuses 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 the 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 around p that contains no points of H, which implies it must contain points from K. Participants suggest exploring a proof by contraposition, considering the implications if p is not a limit point of either set. The discussion emphasizes the necessity of intersections between open intervals and the sets involved to establish the limit point condition. Overall, the proof hinges on the relationships between the open intervals and the elements of the sets H and K.
Jaquis2345
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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?
 
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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