Prove that zero is the only element in this family of nested intervals I'm proving that
$$\bigcap_{n\in\mathbb{Z}_+}\left[-\frac{1}{n},\frac{1}{n}\right]=\{0\}$$
for my real analysis class. I shall not make use of limits since we have not covered them yet. For the proof I use the following theorems (these theorems may be found in Bartle's Introduction to Real Analysis 4th ed. Lemma 2.3.4 and Theorem 2.5.3):
Theorem 1 (approximation property): Let $A\subseteq\mathbb{R}$, $A\ne\varnothing$, and $A$ is bounded. Then a lower bound $b$ for $A$ is the greatest lower bound for $A$, i.e. $b=\inf A$, iff for every $\varepsilon>0$ there exists an $a_\varepsilon\in A$ such that $b+\varepsilon>a_\varepsilon$.
Theorem 2: Let $\mathcal{G}=\left\{[a_n,b_n]:n\in\mathbb{Z}_+, a_n\le a_{n+1},b_n\ge b_{n+1}\right\}$ be a countable family of closed nested intervals, that by the Cantor's Nested Intervals Theorem we know $\cap\mathcal{G}\ne\varnothing$. Thus, if $\inf(\{b_n-a_n:n\in\mathbb{Z}_+\})=0$, then $\cap\mathcal{G}=\{x\}$ for some $x\in\mathbb{R}$, i.e., the set $\cap\mathcal{G}$ has only one element.
Proof: Let $\mathcal{F}=\left\{[-1/n,1/n]:n\in\mathbb{Z}_+\right\}$, and $I_n\in\mathcal{F}$. Since $\mathcal{F}$ is a countable family of closed nested intervals, by Cantor's Nested Interval Theorem we know that $\cap\mathcal{F}\ne\varnothing$. Now we show that $\cap\mathcal{F}$ has only one element.
Consider the set $A=\left\{1/n-(-1/n):n\in\mathbb{Z}_+\right\}$, then
i) Since by definition $2/n>0$ for any integer $n$, then $0$ is a lower bound for $A$.
ii) Let $\varepsilon>0$ be arbitrary. By the Archimedean Property, we know that there exists a positive integer $m$ such that $\varepsilon/2>1/m,$ i.e., $0+\varepsilon>2/m$. Note that $2/m\in A$. Hence, considering that $0$ is a lower bound for $A$ by Theorem 1, we may conclude that $0=\inf A$. Furthermore, by Theorem 2 the previous result implies that $\cap\mathcal{F}$ is a singleton.
Now we show that $0\in\cap\mathcal{F}$. Suppose $0\notin\cap\mathcal{F}$, thus there is an $n\in\mathbb{Z}_+$ such that $0\notin I_n$, that is $0\notin[-1/n,1/n]$, implying that $0<-1/n$ which is impossible, thus $0\in\cap\mathcal{F}$. Hence, $\cap\mathcal{F}=\bigcap_{n\in\mathbb{Z}_+}\left[-1/n,1/n\right]=\left\{ 0 \right\}$.
$\square$
Is this proof correct? Thank you.
 A: A few things:

Since by definition, $\frac{2}{n} > 0$ for any integer $n$.....

This is a minor point but it isn't really true. What you meant to say was that for any positive integer $n$, $\frac{2}{n} > 0$. Also, this typically isn't a statement "by definition". Rather, it's something you can prove from the field/order axioms.

Suppose $0 \notin \bigcap \mathcal{F}........

Right, so you're doing a proof by contradiction here. That's not really necessary, though. See, you already know that:
$$\forall n \in \mathbb{Z}^+: 0 \in \left[-\frac{1}{n},\frac{1}{n} \right]$$
So, it follows, by definition, that $0$ belongs to the intersection of all of these closed intervals. I'm not saying your argument there was wrong or anything; it's just that a contradiction wasn't exactly needed. Of course, it's good that you're trying to be creative and come up with your own proofs.

Now, you needn't have actually used those Cantor's Intersection Theorem for this. You already know that $\{0\} \subseteq \bigcap \mathcal{F}$ by virtue of the last part of what I said above. Let $x \in \bigcap\mathcal{F}$. Then:
$$\forall n \in \mathbb{Z}^+: x \in I_n$$
If $x \neq 0$, then $|x| > 0$. Using the Archimedean property, we can find an $m \in \mathbb{Z}^+$ such that $m > \frac{1}{|x|}$. Or to re-phrase, we can find such an $m$ with the property that $|x| > \frac{1}{m}$. This implies that $x \notin I_m$ and that is impossible. So, it follows that $x = 0$ and we're done.
