Is my proof correct that set $A$ is a subset of $f^{-1}[f[A]]?$

Question

$f: X \to Y$ prove that if $A \subseteq X$, then A is a subset of the pre-image of the image of $A$, which is shown by these symbols respectively: $f^{-1}[f[A]]$

This is my proof:

If $A \nsubseteq f^{-1}[f[A]]$, then there exists "$x$" as an element of $A$, such that $f^{-1}[f[A]] \neq x$ which is a contradiction.

Is this the correct way of going about doing this? I'm also not sure if we can assume that $f^{-1}[f[A]] = x$ is always true

Answer 1

$f^{-1}[f[A]]$ is a set, and $x$ is an element. They cannot be equal.

The correct way of proving this is: let $x\in A$, then $$f(x)\in \{f(x)\ |\ x\in A\}=f[A]$$ by the definition of image. Now because $f(x)\in f[A]$ (and obviously $x\in X$), we have that $$x\in\{x\in X\ |\ f(x)\in f[A]\}=f^{-1}[f[A]]$$ by the definition of preimage. Therefore $A\subseteq f^{-1}[f[A]]$.

Answer 2

You might prove it by contradiction like in your attempt. However $$ A\not\subseteq f^{-1}[f[A]] $$ means that

there exists $x\in A$ such that $x\notin f^{-1}[f[A]]$

There's a big difference between “$\notin$” and “$\ne$”.

Since $x\notin f^{-1}[f[A]]$, we have $f(x)\notin f[A]$. However, as $x\in A$, we have $x\in f[A]$: a contradiction.

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Now we can transform this in a proof without contradiction.

Suppose $x\notin f^{-1}[f[A]]$. Then $f(x)\notin f[A]$ and therefore $x\notin A$.

By contrapositive, if $x\in A$, we must have also $x\in f^{-1}[f[A]]$.

The main fact used is that, for $B$ a subset of $Y$ and $x\in X$, $$ x\in f^{-1}[B]\quad\text{if and only if}\quad f(x)\in B $$ applying this to $B=f[A]$.

Answer 3

It suffices to note that for any $x\in A$ we have some image $f(x)$. This implies $$ x\in f^{-1}(f(x))\subseteq f^{-1}(f(A)) $$ To sum up, we have $$ x\in A\implies x\in f^{-1}(f(A)) $$ which in turn means $A\subseteq f^{-1}(f(A))$.