Continuity- Image of a function

Question

I have the following topological space:

$\tau=$ {$U\subseteq R: 1\notin U$} U {$R$}

and the following application:

$f: (R, \tau)\to (R, \tau)$

I have to see that if f(1)=1, then f is continuous.

Is $f^{-1}(R)=R$??

Answer 1

The preimage of the codomain is always the entire domain, so $f^{-1}[\Bbb R]=\Bbb R\in\tau$.

Your function is continuous since the preimage of a closed set, i.e. of a set containing $1$, always contains $1$. Since the image of set containing $1$ also contains $1$, it is also a closed map.