0
$\begingroup$

Let $f : X \rightarrow Y$ and let $B \subseteq Y$. The inverse image of $B$ is a subset of $X$ defined by $$f^{-1}(B) = \{x \in X | f(x) \in B\}$$ Prove the following claims:

There are several claims that I am required to prove, but I am stuck on the first one, which is:

$$B_1 \subseteq B_2 \Rightarrow f^{-1}(B_1) \subseteq f^{-1}(B_2)$$

So far I have done:

$b \in (B_1 \subseteq B_2) \rightarrow b \in (f^{-1}(B_1) \subseteq f^{-1}(B_2))$

$\forall b [b \in B_1 \rightarrow b \in B_2]$ (Definition of a subset from my class' lecture notes.)

$\rightarrow f^{-1}(b) \subseteq f^{-1}(b)$

I feel like this is the right direction, but that I must have missed a step because it feels too "easy". Any advice?

$\endgroup$
3
$\begingroup$

Think of the function $f$ and the relation $\subseteq$ as 'arrows'. Then you just have to start at the correct place and 'follow the arrows' correctly:

Let $x$ be in $f^{-1}(B_1)$. Then by definition, $f(x)$ is in $B_1$. Since $B_1$ $\subseteq$ $B_2$, $f(x)$ is in $B_2$. Then again by definition, $x$ is in $f^{-1}(B_2)$. QED.

@Austin C : Drat, beat me.

$\endgroup$
  • $\begingroup$ That's okay; your solution is less wordy. $\endgroup$ – 211792 Sep 29 '16 at 17:29
2
$\begingroup$

You want to show that if $x$ is an element of $f^{-1}(B_1)$, then $x$ must also be an element of $f^{-1}(B_2)$ (per the definition of subset). So suppose $x\in f^{-1}(B_1)$. Then there's some $b\in B_1$ so that $f(x)=b$. But since $B_1\subseteq B_2$, you know that $b$ is also an element of $B_2$. That is, we've found an element $b\in B_2$ so that $f(x)=b$. This is precisely the condition for $x$ to be an element of $f^{-1}(B_2)$.


As a separate note, I would encourage you to attempt to write out complete sentences for each step of your argument. The shorthand is very helpful (and indeed I used it in my answer), but when you're first learning the shorthand, writing out sentences can help you root out problems in your argument.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.