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In the definition of a limit point $x$ of a set $A$ ($x$ not necessarily in $A$), it is required that for every $\varepsilon>0$, there exists $y\in A$ such that $$0<|x-y|<\varepsilon$$

Is this a standard definition? Do the inequalities have to be strict? Why is the zero necessary on the left hand side inequality?

user90335
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  • By definition, 0 < $\epsilon$. if 0 was not less than epsilon, then 0 is greater than or equal to epsilon. This is incorrect. – MathApprentice Aug 17 '13 at 00:38
  • Yes, but why require $x\neq y$? – user90335 Aug 17 '13 at 00:42
  • A good answer has been posted, so I've deleted my comment. This is a pretty standard definition. If you are studying metric spaces and perhaps looking toward topology, a limit point $x$ of a set $A$ is sometimes defined as a point such that any neighborhood of $x$ contains a point in $A$. But this definition and your definition are the same if we are looking at the metric space $\mathbb{R}$ with the absolute value as the distance function. – Alex Wertheim Aug 17 '13 at 00:43

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This is the standard definition, and one of the inequalities has to be strict, as I will show: The intuition goes as following: Saying that a point $x$ is a limit point of a set $A$ is the same as saying that "however close I get to $x$, there is always a point of $A$ nearby". So basically, for every small interval around $x$, I want there to be a point of $A$ in it. Using $\epsilon$ notation, for every small positive $\epsilon$, I want there to be a point of $A$ in the interval $(x-\epsilon,x+\epsilon)$. But this is the same as saying that for every $\epsilon>0$, there is a point $y\in A$ that is closer to $x$ than $x-\epsilon$ or $x+\epsilon$; that is, $$x-\epsilon<y<x+\epsilon.$$ By subtracting $x$, we get that $-\epsilon<y-x<\epsilon$, and this is the definition of $|y-x|<\epsilon$.

It is easy to see that you can tweak this a bit and say that we want a point of $A$ in $[x-\epsilon,x+\epsilon]$ for all $\epsilon$, and so we want all $y\in A$ such that $0<|y-x|\leq \epsilon$.

We can't have $|y-x|=0$, however, because that just says that $x=y$, and we want points of $A$ that are close to $x$ but not equal to $x$.

rfauffar
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    Just want to add that $0<$ basically ensures that we don't count isolated points of $A$ as limit points if $x \in A$. If $x \notin A$, then it doesn't matter whether $0<$ is there or not since we can never have $x = y$ for some $y \in A$. – Pratyush Sarkar Aug 17 '13 at 01:44