Proving the Existence of Limit Points through Convergence of Sequences

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To prove 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, one must approach the proof in two directions. First, using the definition of a limit point, for any error 1/n, one can find points y_n in A that are distinct from x and satisfy |y_n - x| < 1/n. As n approaches infinity, these points y_n will converge to x, demonstrating the first direction of the proof. The second direction involves showing that if a sequence converges to x, then x must be a limit point of A. This approach effectively utilizes the definition of convergence and the properties of limit points to establish the equivalence. The discussion highlights the importance of understanding definitions in constructing mathematical proofs.
*melinda*
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The question says:

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

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 x\in A, and would need to construct a sequence that converges to x.
Why are these things always easier said than done :confused: ?

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

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 y_n in A for each n. as n goes to infinity, y_n goes to x since 1/n goes to zero. And you're done. (one way)
 
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