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Sets & limit points and stuff

  • Thread starter *melinda*
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  • #1
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The question says:

Let [itex]A[/itex] be a set and [itex]x[/itex] a number.
Show that [itex]x[/itex] is a limit point of [itex]A[/itex] if and only if there exists a sequence [itex]x_1 , x_2 , ...[/itex] of distinct points in [itex]A[/itex] that converge to [itex]x[/itex].

Now I know from the if and only if statement that I need to prove this thing both ways.

So, the proof in one direction (I think) would be that I have a limit point [itex]x\in A[/itex], and would need to construct a sequence that converges to [itex]x[/itex].
Why are these things always easier said than done :confused: ?

One of the definitions in my book states:
[itex]x[/itex] is a limit point of [itex]A[/itex] if given any error [itex]1/n[/itex] there exists a point [itex]y_n[/itex] of [itex]A[/itex] not equal to [itex]x[/itex] satisfying [itex]|y_n -x|<1/n[/itex] or, equivalently, if every neighborhood of [itex]x[/itex] contains a point of [itex]A[/itex] not equal to [itex]x[/itex].

I feel like I can somehow use this definition, or at least the definition of Cauchy sequences to help with my proof. Only trouble is, I don't know what to do with what I have.

help?
 

Answers and Replies

  • #2
740
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you need the definition of convergence also
 
  • #3
74
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Cauchy is irelevant here. Use your defenition of limit point to select an element [itex] y_n [/itex] in A for each n. as n goes to infinity, [itex] y_n [/itex] goes to x since 1/n goes to zero. And you're done. (one way)
 

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