函数逆像相关命题证明求助：f⁻¹(Y)=X与f⁻¹(Y-B)=X-f⁻¹(B)

Hey there! Let's work through these two inverse image proofs together—they're all about sticking closely to the definitions, which is the key to set theory proofs. First, let's recap the critical definition we'll use for both parts:

For a function $f: X \to Y$ and subset $B \subseteq Y$, the inverse image of $B$ under $f$ is defined as:
$$f^{-1}(B) = { x \in X \mid f(x) \in B }$$

a) Prove $f^{-1}(Y) = X$ #

We'll show equality by proving both subset inclusions (this is the standard way to prove set equality):

1. $X \subseteq f^{-1}(Y)$:
   Take any arbitrary $x \in X$. Since $f$ is a function from $X$ to $Y$, every element of $X$ must map to some element in $Y$. That means $f(x) \in Y$. By the definition of inverse image, this directly implies $x \in f^{-1}(Y)$. Since this holds for every $x$ in $X$, $X$ is a subset of $f^{-1}(Y)$.

2. $f^{-1}(Y) \subseteq X$:
   By the very definition of inverse image, every element in $f^{-1}(Y)$ is an element of $X$ (we only consider $x \in X$ when defining $f^{-1}(B)$). So trivially, $f^{-1}(Y)$ is a subset of $X$.

Since both subsets contain each other, we can conclude $f^{-1}(Y) = X$.

b) Prove $f^{-1}(Y - B) = X - f^{-1}(B)$ #

Again, we'll prove mutual subset inclusion to establish equality:

Step 1: $f^{-1}(Y - B) \subseteq X - f^{-1}(B)$ #

Let $x \in f^{-1}(Y - B)$. By the inverse image definition:
$$f(x) \in Y - B$$
This means two things:

• $f(x) \in Y$ (which is always true for any $x \in X$, since $f$ maps to $Y$)
• $f(x) \notin B$

If $f(x) \notin B$, then $x$ can't be in $f^{-1}(B)$ (because $f^{-1}(B)$ only includes $x$ where $f(x) \in B$). Since $x$ is in $X$ (from the inverse image definition), this means $x \in X - f^{-1}(B)$.

Step 2: $X - f^{-1}(B) \subseteq f^{-1}(Y - B)$ #

Let $x \in X - f^{-1}(B)$. This implies:

• $x \in X$
• $x \notin f^{-1}(B)$

The second condition translates to $f(x) \notin B$ (by the inverse image definition). But since $x \in X$, $f(x)$ must be in $Y$. Combining these, $f(x) \in Y$ and $f(x) \notin B$, so $f(x) \in Y - B$. By the inverse image definition, this means $x \in f^{-1}(Y - B)$.

Since both directions hold, the sets are equal: $f^{-1}(Y - B) = X - f^{-1}(B)$.

内容的提问来源于stack exchange，提问作者Luis Robles