# Set theory proof on injections, sequences of sets

I'm trying to practice my proof-writing skills. At the moment I trying to prove the following proposition:

For every one-to-one function $\: f:X \to Y \:$, and sequence of sets $\: B_n \subseteq X \:$,

$$\: f \left[\bigcap_{n=0}^\infty B_n \right] = \bigcap_{n=0}^\infty f \left[B_n\right].$$

I was hoping to get some feedback on the proof. I suppose both in terms of content and form. Thanks!

Proof.

First, for some arbitrary $y \in Y$, suppose that $y \in f\left[ \bigcap_{n=0}^\infty B_n\right]$.

Then, by definition, $\exists x \in \bigcap_{n=0}^\infty B_n$ such that $f(x) = y$.

And, because we have an intersection, it follows that $\exists x : \forall n\, (x \in B_n)$.

In that case, $y \in f\left[B_n\right]$ for each $n$.

And therefore, $y \in \bigcap_{n=0}^\infty f\left[B_n\right]$.

So we've shown that $y \in f\left[ \bigcap_{n=0}^\infty B_n\right] \implies y \in \bigcap_{n=0}^\infty f\left[B_n\right]$.

Now, assuming that $f$ is one-to-one, suppose that we have $y \in f\left[B_n\right]$ for each $n$.

So for all $n$, there is an $x \in X$ such that $f(x) = y$.

But if $f$ is an injection then, by definition, there is only one such $x$.

It follows that for each $n$, this unique $x$ is in $B_n$, that is $x \in \bigcap_{n=0}^\infty B_n$.

And $x \in \bigcap_{n=0}^\infty B_n \implies y \in f\left[ \bigcap_{n=0}^\infty B_n\right]$.

So we've shown that $y \in \bigcap_{n=0}^\infty f\left[B_n\right] \implies y \in f\left[ \bigcap_{n=0}^\infty B_n\right]$.

So the implication goes in both directions, which is what we needed to show.

End proof.

@WilloW In the first inclusion you took $y\in Y$ first and you supposed it was in LHS set of the equality. Despite not being incorrect, there's no need for that. You can simply take $y$ in the LHS set from the get go. –  Git Gud Feb 2 '13 at 23:24