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Alright, I posted this earlier and didn't really provide any context or format it correctly, so I figured I would fix all of that up. I'm a little confused on the idea of a limit point, so I was hoping to get some clarification. Our definition of a limit point is as follows:

We say that a point $p$ is a limit point of X if every neighborhood of $p$ contains infinitely many elements of X.

Equivalently, we say that $p$ is a limit point of X if every neighborhood of $p$ contains at least one point of X which is different from $p$.

I'm confused about where to go from here using our definition of a limit point.

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    I suppose you wanted to say $0<|p-y|<\varepsilon$ in the title. – 2017-02-13
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    Our homework says x, not $p$. – 2017-02-13
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    But you didn't define $x$. – 2017-02-13
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    Yeah...neither did my homework...not sure what that is about. – 2017-02-13
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    I suppose $x$ should be replaced with $p$, as it's the only way it make sense. – 2017-02-13
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    I guess I'll email my professor. – 2017-02-13
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    How did your professor defined neighborhood of $p$? – 2017-02-13
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    My professor said it was a typo and should've been $p$. We define an open neighborhood as "Let x be in R. An open interval which contains x is called a neighborhood of x." Also, I'm trying to format correctly and i have no clue what i'm doing. I tried doing the "an element of" sign, but it just shows up as \in. – 2017-02-13
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    I posted an answer. Just ask if you don't understand. – 2017-02-13
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    I understood it completely. Thank you! – 2017-02-13
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    I didn't realize I had the reputation to do so yet. Thanks. – 2017-02-14
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    Yeah I completely understand. Thanks! – 2017-02-14

2 Answers 2

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Suppose $p$ is a limit point. We want to prove that for all $\varepsilon>0$, there exists $y$ such that $|y-p|<\varepsilon$. Let us take $\varepsilon>0$. As $(p-\varepsilon, p+\varepsilon)$ is a neighbourhood of $p$, there exists $y\in A$ that satifies $y\in (p-\varepsilon, p+\varepsilon)$. That is: $$y\in(p-\varepsilon, p+\varepsilon) \iff p-\varepsilon < y < p +\varepsilon \iff -\varepsilon < y-p < \varepsilon \iff |y-p|<\varepsilon$$

so $y\in A$ and $|y-p|<\varepsilon$.

Now, suppose that for all $\varepsilon>0$, there exists a point $y\in A$ such that $|p-y|<\varepsilon$ (that means $y\in(p-\varepsilon, p+\varepsilon)$.) We want to prove that $p$ is a limit point for $A$: that is, every neighbourhood of $p$ contains a point of $A$. Let $V$ be a neighbourhood of $p$, it must contain an interval of the form $(p-\varepsilon, p+\varepsilon)$ for a certain $\varepsilon$. So there is an $y\in A$ that satisfies $y\in(p-\varepsilon, p+\varepsilon)$, because of our hypothesis. We conclude noting that $(p-\varepsilon, p+\varepsilon)\subset V$, so that element $y\in A$ also satisfies $y\in V$.

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    Okay. I'm confused about this. It's an "if and only if" statement so i'm proving both ways. I just don't get which way you're proving. I don't see where limit points come into your proof at all. – 2017-02-13
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    Okay, I'm adding more details. I'm actually proving both. – 2017-02-13
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    Thank you. I'm really trying to understand what you're proving. The idea of limit points is confusing me. – 2017-02-13
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    No problem, I only wrote the general lines for it. Please ask anything you don't understand. – 2017-02-13
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    Okay. How do you go from $y\in $A to $|y-p|$ < e? Actually, just kidding. I think I understand that. You know its in the neighborhood so you're just rearranging the inequality and then taking absolute value. I got that. Let me keep reading. – 2017-02-13
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    I got it. I think I understand that part. Unfortunately, this question has 3 more parts so I'll try to do those as well. Thanks a lot. – 2017-02-13
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    You're welcome. – 2017-02-13
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53591/discussion-between-johnie4usc-and-flytothesurface). – 2017-02-14
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$\Rightarrow$ Assume that $p\in\Bbb R$ is a limit point of $A$. Let $\epsilon>0$. Then $(p-\epsilon,p+\epsilon)$ is a neighborhood of $p$. Thus, $(p-\epsilon,p+\epsilon)$ contains a point of $A$ different from $p$. This means that there exists $y\in A$ such that $y\neq p$ and $y\in (p-\epsilon,p+\epsilon)$. Now, $$\begin{align} y\in(p-\epsilon,p+\epsilon)&\iff p-\epsilon

$\Leftarrow$ Let $p\in\Bbb R$. Assume that for every $\epsilon>0$ there exists $y\in A$ such that $0<|y-p|<\epsilon$. Let $I$ be a neighborhood of $p$. Then $I$ is an open interval that contains $p$. Write $I=(a,b)$. Take $$\epsilon=\min\{p-a,b-p\}.$$ Then $\epsilon>0$ and $(p-\epsilon,p+\epsilon)\subset I$. Using the hypothesis,there exists $y\in A$ such that $0<|y-p|<\epsilon$. This means that $y\neq p$ and $y\in(p-\epsilon,p+\epsilon)$. Clearly, $I$ contains $y$ and $y$ is different from $p$. Thus, $p$ is a limit point of $A$.