TITLE: Show that $0$ is a limit point of $L^-$
QUESTION [0 upvotes]: I need to show that $0$ is a limit point of $L^-$, the set of negative numbers in a real number line.
Given the definition:

If $M$ is a point set and $x$ is a point, then $x$ is a limit point of $M$ if every open interval containing $x$ contains a point of $M$ different from $x$.

I'm really lost on this. The best I can figure is that $M=(a,b)$ and $a<0<b$, but that really doesn't help explain how $0$ is a limit point in $L^-$. If $L^-$ is a set of negative numbers, how can $0$ be a limit point? In the interval $(a,b)$, wouldn't b always be positive if the interval contains $0$?

REPLY [0 votes]: We have $L^-$ which is $x\in\mathbb{R},x\lt0$. $L^-$ translates to $M$ in your definition. Now any open interval that contains $0$ must contain an element of $L^-$ that is not $0$. Hence $0$ is a limit point of $L^-%$.
{"set_name": "stack_exchange", "score": 0, "question_id": 1436446}