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?
 A: 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.
A: Let $x$ be in the closure of $C$.
There exists $x_n \in C$ such that $|x_n - x| < 1/n$ for every integer $n > 0$
If $m > n$, then $|x_m - x| < 1/m < 1/n$.
Hence $\lim_n x_n = x$.
By the assumption, $x \in C$.
Hence $C$ is closed.
A: What you seem to be asking is this: If $C$ is a subset of $\mathbb{R}$, such that for any convergent sequence of elements of $C$, the limit is also an element of $C$, then must $C$ be closed?
The answer is Yes; topological spaces that satisfy this condition are called "sequential spaces," and $\mathbb{R}$ is a sequential space.  Here's why:
Suppose that $x\in\mathbb{R}$, and that every neighborhood of $x$ intersects $C$.  Then for each natural number $n$, let $x_n$ be a point in the intersection of $C$ with $(x-\frac 1n, x+\frac 1n)$.  Then $x_n\to x$, so $x\in C$.  Therefore $C$ must be closed.
This argument works whenever you replace $\mathbb{R}$ with a "first-countable" topological space (each point admitting a countable neighborhood basis, here given by $\{(x- \frac 1n, x+\frac 1n): n\in\mathbb{N}\}$).  There are, however, sequential spaces which are not first-countable.  
