Proving the Existence of Limit Points through Convergence of Sequences

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SUMMARY

The discussion centers on proving that a number x is a limit point of a set A if and only if there exists a sequence of distinct points in A that converges to x. The proof requires demonstrating both directions of the statement. The definition of a limit point is utilized, stating that for any error 1/n, there exists a point y_n in A, distinct from x, such that |y_n - x| < 1/n. By selecting appropriate elements from A, the proof shows that as n approaches infinity, y_n converges to x, thereby establishing the required relationship.

PREREQUISITES
  • Understanding of limit points in topology
  • Knowledge of convergence of sequences
  • Familiarity with the epsilon-delta definition of limits
  • Basic concepts of neighborhoods in metric spaces
NEXT STEPS
  • Study the epsilon-delta definition of limits in depth
  • Explore the properties of convergent sequences in real analysis
  • Learn about the concept of neighborhoods in metric spaces
  • Investigate the role of distinct sequences in proving limit points
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Mathematics students, particularly those studying real analysis or topology, as well as educators seeking to clarify the concept of limit points and convergence of sequences.

*melinda*
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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?
 
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you need the definition of convergence also
 
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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