# Thread: Real Analysis: Limit Point

1. ## Real Analysis: Limit Point

Q: Suppose that A and B are non-empty sets of real numbers and that x is a limit point of AUB. Prove that X is a limit point of A or of B.

my solution:

Given x is a limit point of AUB

then for all epsilon > 0, nbd(x) contains a point of AUB different from x

Then nbd(x) contains a point of A or nbd(x) contains a point of B different from x.

Therefore x is a limit point of A or of B.

[nbd = neighborhood]

Just wondering if my approach is correct or not. I feel it's not as easy as I thought...

2. You are correct, that is not sufficient to prove it.
Suppose that x is not a limit point of A.
Then some neighborhood, $N(x)$ contains no point of $A-\{x\}$.

But for any neighborhood, $M(x)$ then $Q(x)=M(x)
\cap N(x)$
is also a neighborhood of $x$ which must contain points of $(A\cup B)}-\{x\}$.

Does that make $x$ a limit point of $B?$