To prove that $ \cap_{n=1}^{\infty} (0 , \frac{1}{n}) = \emptyset $

Question

$$ \bigcap_{n=1}^{\infty} \left(0 , \frac{1}{n}\right) = \varnothing$$

Now assume that intersection contains $b$. So, $ b < \frac{1}{n} \forall n \in N$. Since $b > 0$, so we have by Archimedian property that $\exists n \in \mathbb{N}$ such that $ \frac{1}{n} < b$ which is a contradiction to assumption that $b < \frac{1}{n}$

Is this correct ?

Thanks

Answer 1

Looks good to me.

Small notational nitpicks: It's \Bbb N ("BlackBoard Bold") to make $\Bbb N$.

And when using symbols to convey logic, like in your $b<\frac1n\forall n\in \Bbb N$, the quantifiers ($\forall$ and $\exists$) always come before whatever it is they modify.

So while one in English could say "$b$ is smaller than $\frac1n$ for any natural number $n$", the symbolic statement must be $\forall n\in \Bbb N, b<\frac1n$ (exactly how to separate $\forall n\in \Bbb N$ and $b<\frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<\frac1n$ in parentheses).