In Class Problems number 2

Prop 2. Suppose that $E\subset\mathbb{R}$ is closed and that $\left(x_{n}\right)$ is a convergent sequence of numbers in $E$. Then
lim $\left(x_{n}\right)\in E$.

Proof. let $\left(x_{n}\right)_{n = 1}^\infty \rightarrow x$ and by contradiction suppose $x\notin E$. It follows that $x\in \mathbb{R} - \{E\} = E^{c}$. Note that $E^{c}$ is open. So $\exists$ $r > 0 \ni B_{r}(x) \subseteq E^c$. This contradicts the fact that $\left(x_{n}\right)_{n = 1}^\infty \rightarrow x$, Therefore $x\in E$. $\blacksquare$

I agree with you until that last step there. It feels a bit circular and imprecise to me. I would say the reason this is a contradiction, is because there is a $B_r(x)\subset E^c$. By the definition of a limit, we can get $x_n$ within $\epsilon$ of $x$ for any $\epsilon>0$. However, we cannot do this for $\epsilon < r$, because the terms $x_n\in E$, while $B_r(x)\subset E^c$.
Another interesting counter example I was thinking of was $(x_n)_{n=1}^{\infty}=\frac{1}{n}$ over $\mathbb{Q} -\{0 \}$. The integers are closed, and the sequence obviously converges to $0$, but the limit lies outside of the set where the terms live. Is this example false because the set is no longer closed when we remove one point? Or some other reason? I had trouble understanding what the problem would be with this situation.

@shill2 We can make the ball of radius $r$ as small as we like about $x$ and it will always be in $E^c$. The definition of the limit says that the terms of the sequence can get as close as we like to $x$, but never reach it. all the $x_{n}$ live in $E$ so they cannot get arbitrarily close to $x$, they can only get within at most $r$ of $x$. Thus the definition of the limit is contradicted.

Exactly. I wasn't disagreeing, only formally defining that we cannot get as close as we like (ie we can't have $\epsilon < r$). Do you have a solution for that counter example though?

@shill2 $(\frac{1}{n})_{n= 1}^{\infty}$ is not defined over $\mathbb{Q} - \{0\}$ so that's not a counterexample. Though $\mathbb{Q} - \{0\}$ is open (EDIT: see below post)

How can $\mathbb{Q}-\{0\}$ be closed. it has sequences that converges to 0

$\mathbb{Q}$ itself is not closed, so subtracting a single point won't help matters.

@nklausen @Cromer You're right I was derping and thinking $\mathbb{Q}$ was closed.

You are correct. Thanks for the correction.