topology question: Define $\tau_f= \{f^{-1}(V):V\in \tau\}$ Suppose $(Y, \tau)$ is a topological space and $f:X\to Y$ is a mapping. Define $\tau_f= \{f^{-1}(V):V\in \tau\}$.
Part a: Show that $\tau_f$ is a topology in X.
My attempt:
First, I need to show $\emptyset \in \tau_f$ and $X \in \tau_f$. Reason: Let $V = \emptyset \in \tau$, then $\tau_f=f^{-1}(\emptyset)=\emptyset$.
Similarly, Let $V = Y \in \tau$, then $\tau_f=f^{-1}(Y)=X$.
Second, I need to show if $f^{-1}(V_i)\in \tau_f$ for $i=1,...,n$, then $f^{-1}(V_1) \cap f^{-1}(V_2) \cdot \cdot \cdot f^{-1}(V_n) \in \tau_f$.
Reason: $f^{-1}(V_i)\in \tau_f$ implies that $V_i \in \tau$;
$f^{-1}(V_1) \cap f^{-1}(V_2) \cdot \cdot \cdot f^{-1}(V_n) = f^{-1}(V_1 \cap V_2 \cap \cdot \cdot \cdot \cap V_n)= f^{-1}(\text{a set} \in \tau) \in \tau_f$.
Third: I need to show if $\{ f^{-1}(V_\alpha) \}$ is an arbitrary collection of member of $\tau_f$, then $\cup_\alpha f^{-1}(V_\alpha) \in \tau$.
Reason: $\cup_\alpha f^{-1}(V_\alpha) = f^{-1} (\cup_\alpha  V_\alpha) = f^{-1}(\text{a set} \in \tau) \in \tau_f$.
Is my solution above correct?
Part b: Suppose $Y=\mathbb{R}$ with the usual topology, that is, $\tau$ is the collection of all open sets of $\mathbb{R}$ in the usual sense, $X$ is the set of all real numbers, and $f: X \to Y$ is defined to be $f(x)=0$ if $x<0$ and $f(x)=1$ if $x\geq 0$.
What is $\tau_f= \{f^{-1}(V):V\in \tau\}$? (That is, characterize or write out all the members of $\tau_f$)
My attempt: Of course $\tau_f$ can be $\mathbb{R}$ and $\emptyset$. Example: $f^{-1}(-3,-1) = \emptyset$, $f^{-1}(-1,2) = \mathbb{R}$.
Also, $f^{-1}(0.5,1.5) = [0,\infty)$; $f^{-1}(-1,0.5) = (-\infty, 0)$.
Are these 4 cases all possible cases? I am not sure.
 A: Your part (a) looks good (the key being that preimage commutes with union and intersection).

For (b): you should try to prove that these are the only 4 possible cases. The examples you gave are good, so let's try to see why those examples give the different cases (i.e. what are the properties of these examples that are important here). For any open set $V\subseteq\mathbb R$ (i.e. $V\in\tau$), I claim that the following are the important properties that generalize your examples:
$$\begin{align}&(1)\quad 0,1\notin V \\&(2)\quad 0,1\in V \\&(3)\quad 1\in V \text{ but } 0\notin V \\&(4)\quad 0\in V \text{ but } 1\notin V.\end{align}$$
What do you get for $f^{-1}(V)$ in each case? Are there any other cases to consider?
A: $
\newcommand{\II}{\mathcal{I}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\invf}[1]{f^{-1} \left( #1 \right)}
$Your proof for first part's bits about intersection and union could use a little clarifying. You simply need to use the properties that
$$\bigcup_{i \in \II} \invf{U_i} = \invf{ \bigcup_{i \in \II} U_i } \qquad \bigcap_{i \in \II} \invf{U_i} = \invf{ \bigcap_{i \in \II} U_i }$$
for any indexing set $\II$. These are purely set-theoretic properties, for clarity.

For the second part, note that a given open set $U$ in $\R$ will satisfy one of the following:

*

*$0,1 \not \in U$

*$0 \not \in U, 1 \in U$

*$1 \not \in U, 0 \in U$

*$0,1 \in U$
Hence those indeed are the only four cases to worry about.
