# Baby Rudin Theorem $4.14$ [closed]

In Baby Rudin theorem $$4.14$$, he says:

$$f(f^{-1}(E))\subset E \quad\forall E \subset Y$$ and then

$$f^{-1}(f(E)) \supset E \quad$$ if $$\quad E \subset X$$.

I thought functions were invertible $$\iff$$ bijective (https://en.wikipedia.org/wiki/Bijection). How would that not imply set equality in each case? Your help would be appreciated.

• $f^{-1}(E)$ is defined as $\{x: f(x) \in E\}$ and this makes sense whether or not $f$ is invertible. Apr 12 at 12:30
• Ah so $f^{-1}(E)$ is not really a function in some sense? Apr 12 at 12:42
• @Rudinable $f^{-1}(E)$ is a set :) Apr 12 at 12:47
• $f^{-1}(E)$ is the preimage. It is an unfortunate use of notation because preimages and inverse functions has little relation to each other. Apr 12 at 13:37

Given a function $$f\colon X\to Y$$, we get two associated functions:
1. The direct image function $$\underline{f}\colon\mathscr{P}(X)\to\mathscr{P}(Y)$$, (where $$\mathscr{P}(A)$$ is the powerset of $$A$$, the set whose elements are all subsets of $$A$$), defined by letting $$\underline{f}(A)=\{f(a)\mid a\in A\}$$ for any $$A\subseteq X$$; and
2. The inverse image function $$\overline{f}\colon \mathscr{P}(Y)\to\mathscr{P}(X)$$, given by $$f^{-1}(B) = \{x\in X\mid f(x)\in B\}$$ for any subset $$B$$ of $$Y$$.
By abuse of notation, we usually denote $$\underline{f}$$ by $$f$$, and $$\overline{f}$$ by $$f^{-1}$$; there is usually no danger of confusion, because we are normally dealing with sets in which there is no subset of $$X$$ which is also an element of $$X$$, and no subset of $$Y$$ is also an element of $$Y$$. So if we write $$f(A)$$, we will know if $$A$$ is a subset or an element of $$X$$, and it will normally not be both, so we know whether we mean the original function or the direct image function; and if we write $$f^{-1}(E)$$, then we know whether $$E$$ is a subset of $$Y$$ (in which case we are talking about the inverse image function), or it is an element of $$Y$$, in which case "$$f^{-1}(E)$$" would have to refer to the inverse function of $$f$$ which, as you note, only exists when $$f$$ is bijective.
Here, Rudin is working with the inverse image and direct image functions; you can tell because the argument $$E$$ is given as a subset of $$Y$$ (and later of $$X$$), rather than as an element.