Set $E = \mathbb R \setminus \{\frac{1}{n} | n \in \mathbb N \}$ is Open in $\mathbb R$ ? Is it Closed in $\mathbb R$?
We can find it directly from taking complement of set. Complement would be $ E' = \{\frac{1}{n} | n \in \mathbb N \} $
Complement would not be Open because there won't be any interior points. Complement would not be Closed because it doesn't contain limit point $0$. So we can say that set $E$ is closed and open from complement method.
Now check it without taking complement.
Set E won't be open because $0$ won't be interior point.
Set E won't be closed because it doesn't contain many limit points. E.g. $1$ is limit point of sequence $\{ 1 + \frac{1}{n}$ $ | $ $ n \in \mathbb N \}$.
Why I am getting different answers ?
Edit: I interpreted theorem incorrectly. Theorem is:
Set is closed if and only if its complement is open.
So above complement is neither closed nor open, we can't say that set $E$ is open and closed.
