Determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open I want to determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open. Here is my attempt at the problem, but I am having some trouble with this idea so please point out any errors. Thank you!
Claim: ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open.
Let $x \in {\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$. By the definition of an intersection we have that $x$ is an element of every set. That is,
\begin{equation*}
0 < x < \frac{1}{n},  \forall n\in\mathbb{N}^+.
\end{equation*}
By the density of the rationals we have that there exists $r_1,q_{n}\in\mathbb{Q}$ such that
\begin{equation*}
0 < r_1 < x < q_{n} < \frac{1}{n}, \forall n\in\mathbb{N}^+.
\end{equation*}
That is, for every $x\in A$ there exists $h > 0$ such that
\begin{equation*}
x - h = r_1
\end{equation*}
and
\begin{equation*}
x + h = q_{n}
\end{equation*}
so we have that
\begin{equation*}
(x - h, x + h) = (r_1, q_{n}) \subset (0, \frac{1}{n}).
\end{equation*}
The set is open.
 A: The set is open. Also closed. Indeed, it is the empty set.
A: For proving that a set $X$ is open, you need to show $\forall x\in X,\exists p,q$ such that $x\in(p,q)\subseteq X$, where $p,q$ are constants.
In your construction, $r_1,r_2$ depend $n$. A choice of $r_1,r_2$ that works for $n_1$ may not work for $n_2$.
It is true that $\forall x\in(0,1/n),x\in(r_1,r_2)\subseteq(0,1/n)$, which indicates that $(0,1/n)$ is open. But for given $r_1,r_2$, the set $(r_1,r_2)\not\subseteq\cap_{n\in\Bbb N}(0,1/n)$ as $\exists m\in\Bbb N$ such that $1/m<r_1$ so $(r_1,r_2)\not\subseteq(0,1/m)$.
In fact you can prove that your set is empty. For every $x>0,\exists n\in\Bbb N$ such that $1/n<x$ i.e. $x\notin(0,1/n)$ so $x\notin\cap_{p\in\Bbb N}(0,1/p)$.
A: You can easily see that the given intersection is empty. Suppose that it isn’t. Then, there is an $x$ such that:
$$\forall n \in \mathbb{N}: x \in (0,\frac{1}{n})$$
However, we know, by the Archimedean Property, that:
$$\exists N \in \mathbb{N}: N > \frac{1}{x}$$
That is, $x > \frac{1}{N}$. So, $x \notin (1,\frac{1}{N})$ and that is impossible. Hence, the given intersection is empty. So, it is, indeed, open.
With regards to your specific argument, I can definitely say that it’s vacuously correct. On the other hand, you’ve introduced an “$A$” in there without explaining what that is. Moreover, we’re led to believe that there is a $h > 0$ such that $x+h$ and $x-h$ are both rational. This doesn’t have to be true because it would imply that $x$ is rational and you haven’t explained why that has to be true.
The gist of your argument (showing that every point in the given set has an open ball centered on it that is a subset of the given set) makes sense and is a good way to approach it. You should reformulate your argument so it’s more precise.
