Topology and continuity I have a function$$ f : (\mathbb{R}, O) \longrightarrow (\mathbb{R}, \text{standard})$$ be given by $f(x) = x.$ I am new to this so based on the different topology in general how to determine and show if $f$ continuous or not given:
• $O = \{∅, \mathbb{R}\}$;
• $O = \text{standard}$;
• $O$ is the discrete topology on $\mathbb{R}$?
 A: Directly from the definition: $f:X\to Y$ is continuous iff $f^{-1}(U)\subseteq X$ is open for open $U\subseteq Y$.
Now since your function $f$ is special, namely $X=Y=\mathbb{R}$ and $f(x)=x$ then for any subset $A\subseteq\mathbb{R}$ we have $f^{-1}(A)=A$.  Therefore we can reduce the problem to:

Statement: the standard topology $\tau$ on $\mathbb{R}$ is a sub-topology of $O$, i.e. $\tau\subseteq O$.



*

*$O=\{\emptyset, \mathbb{R}\}$ - no, because $(0,1)$ is open in $\mathbb{R}$ while $(0,1)\not\in O$.

*Obviously if $O$ is standard then they are equal, in particular the statement holds.

*By the definition the discrete topology consists of all subsets of $\mathbb{R}$. In particular it contains the standard topology, hence the statement is true.

A: *

*$f$ is not continuous function on $\mathbb{R}.$ Take $U=(0,1) \in (\mathbb{R}, standard),$ and you can check that $f^{-1}(U)$ is not open in $(\mathbb{R}, indiscrete).$

*$f$ is continuous on $\mathbb{R}$

*$f$ is continuous on $\mathbb{R}$
