A question on open and closed sets Suppose $C$ is a subset of $\mathbb{R}$ and for any sequence of points $(x_n)$ in $C$ so that $(x_n)$ converges. Suppose $\lim(x_n) = x$ which is an element of $C$. Is $C$ closed? Why or why not?
I would think it has to be closed if $(x_n)$ converges correct?

If all convergent sequences of points of $C$ have their limits in $C$, then yes, $C$ must be closed. You can prove it by showing that $\Bbb R\setminus C$ is open. If $\Bbb R\setminus C$ is not open, there is a point $x\in\Bbb R\setminus C$ such that for every $\epsilon>0$, $(x-\epsilon,x+\epsilon)\nsubseteq\Bbb R\setminus C$, or in other words, $(x-\epsilon,x+\epsilon)\cap C\ne\varnothing$. Thus, for each $n\in\Bbb Z^+$ there is a point
$$x_n\in\left(x-\frac1n,x+\frac1n\right)\cap C\;.$$
Can you show that $\langle x_n:n\in\Bbb Z^+\rangle$ is a convergent sequence of points of $C$ whose limit is $x$? Your hypothesis would then imply that $x\in C$, contradicting our choice of $x$ and proving that $\Bbb R\setminus C$ must be open, so that $C$ must be closed.
