# Thread: Density of a Set

1. ## Density of a Set

Let $\displaystyle Y\subset X$ and let $\displaystyle A\subset X$. If $\displaystyle A$ is countable, $\displaystyle X$ is complete, and $\displaystyle Y\cup A=X$, is $\displaystyle Y$ necessarily dense in $\displaystyle X$?

Prove it or give a counterexample.

If the answer is no, then does making $\displaystyle X$ compact change anything?

2. Originally Posted by redsoxfan325
Let $\displaystyle Y\subset X$ and let $\displaystyle A\subset X$. If $\displaystyle A$ is countable, $\displaystyle X$ is complete, and $\displaystyle Y\cup A=X$, is $\displaystyle Y$ necessarily dense in $\displaystyle X$?

Prove it or give a counterexample.

If the answer is no, then does making $\displaystyle X$ compact change anything?
Kind of an obvious counterexample, isn't there? Take A and Y to be completely disjoint!

3. Originally Posted by HallsofIvy
Kind of an obvious counterexample, isn't there? Take A and Y to be completely disjoint!
But $\displaystyle X$ is complete. I feel like that can't happen if $\displaystyle Y$ and $\displaystyle A$ are disjoint.

4. $\displaystyle X = \left[ { - 1,0} \right] \cup \left\{ {n^{ - 1} :n \in \mathbb{Z}^ + } \right\}$?

5. Originally Posted by Plato
$\displaystyle X = \left[ { - 1,0} \right] \cup \left\{ {n^{ - 1} :n \in \mathbb{Z}^ + } \right\}$?
I think that works. It would be complete because $\displaystyle Y$ is complete and the only Cauchy sequences in $\displaystyle A$ are all the same number or the sequence $\displaystyle \{a_n\}=\frac{1}{n}$ and both of those converge to something in $\displaystyle X$.

It also answers the second part of the question because this set is also compact.