Let B ⊂ R. Let L be the set of all limit points of B. Prove that B ∪ L contains all its limit points. Let B ⊂ R. Let L be the set of all limit points of B. Prove that B ∪ L contains all its limit points.
I have a question about this one.
Here is my proof
Suppose x is a limit point of B ∪ L and x does not belong to B ∪ L
If x is a limit point of B ∪ L, then x is either limit point of B or limit point of L (or both)
By assuming that x does not belong to B ∪ L
x is not in B and x is not in L
Here is where I am stuck and not sure whether this is right approach
Can anyone give me some ideas?
 A: Suppose that $w$ is a limit point of $B\cup L$, then for all open ball $\beta_r(w)$ there is a $x\in \beta_r(w)$ such that $x\in B$ or $x\in L$, If $x\in L$ then there is $t<r$ such that $\beta_t(x)\subset\beta_r(w)$, how $x\in L$ then there is a $y\in\beta_t(x)$ such that $y \in B$ because $x$ is a limit point of $B$, therefor $w$ is a limit point of $B$ then $w\in L$ then $w\in B\cup L$.
A: An alternative proof (not entirely by contradiction). Let $x$ be a limit point of the union of $B$ and $L$. 
Either every neighbourhood of $x$ contains a point of $B$ or every neighbourhood of $x$ contains a point of $L$, because otherwise there is a neighbourhood $V$ of $x$ with no points from $B$ and another neighbourhood $W$ with no points from $L$. If that were the case, $V\cap W$ would be a neighbourhood of $x$ with no points from the union, contradicting the definition of $x$. This proves the second statement in your proof.
Now,


*

*The first case from the last paragraph means that $x$ is a limit point of $B$, so it belongs to $L$.

*In the second case, $x$ is a limit point of $L$. I claim that it also is a limit point of $B$. Indeed, any neighbourhood $U$ of $x$ contains a point in $L$, call it $y$. Note that $U$ is also a neighbourhood of $y$, and since $y$ is a limit point of $B$, it must contain a point from $B$ too. This means that $x$ is also limit point of $B$. 
To sum up, in both cases $x$ is a limit point of $B$, and therefore belongs to the union. The result follows.
A: "If x is a limit point of B ∪ L, then x is either limit point of B or limit point of L (or both)"
I don't think you can state this without proving it.  A limit point, p, of B $\cup$ U is one in which every neighborhood has a point other than p that is in B $\cup$ U.  I can imagine, however, that one neighborhood has a point in B while another has a point in U but not every neighborhood consistently has a point in either one.
But if p is a limit point of B $\cup$ L then every neighborhood N has a point q $\in$ B $\cup$ L.  If q is in L then q is a limit point of B so every of its neighborhoods has a point r in B.  Choose a neighborhood of q that has a small enough to be a subset of N.  So r $\in$ N.  So every neighborhood of q has a point in B.  So if p is limit point of B $\cup$ L the p is a limit point of B and thus a member of L.  So B $\cup$ L cointains all its limit points.
A: You should prove a bit more rigorously your claim that "$x$ being a limit point of $B\cup L$ means $x$ is either a limit point of $B$ or a limit point of $L$". Once you can state that for certain, the next step I'd take is to say that $x$ cannot be a limit point of $B$, as that would mean $x \in L \subset B\cup L$. This forces $x$ to be strictly a limit point of $L$. Can you proceed from here to a contradiction?
