0
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ $f^{-1}(E)$ is defined as $\{x: f(x) \in E\}$ and this makes sense whether or not $f$ is invertible. $\endgroup$ Apr 12 at 12:30
  • $\begingroup$ Ah so $f^{-1}(E)$ is not really a function in some sense? $\endgroup$
    – Rudinable
    Apr 12 at 12:42
  • $\begingroup$ @Rudinable $f^{-1}(E)$ is a set :) $\endgroup$
    – Oscar
    Apr 12 at 12:47
  • 1
    $\begingroup$ $f^{-1}(E)$ is the preimage. It is an unfortunate use of notation because preimages and inverse functions has little relation to each other. $\endgroup$ Apr 12 at 13:37

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .