0
$\begingroup$

There is very well known lemma: Let $\mathit f$: $X \mapsto Y $ be a mapping.

Let $\mathcal G \subseteq \mathcal P(Y)$ be a collection of subsets of $Y$. $\\$

then $\mathit f^{-1}(\sigma(\mathcal G)) = \sigma(\mathit f^{-1}(\mathcal G))$. $\\$

I do not understand this lemma. The domain of the function $\mathit f$ is $X$, and the codomain is $Y$. $\\$

Thus, the domain of the function $\mathit f^{-1}$ is $Y$, not $2^{Y}$. The function $\mathit f$ is just not defined on $2^{Y}$. But obviously the elements of $\mathcal G$ and $\sigma(\mathcal G)$ consist of the elements of $2^{Y}$? Where is the mistake in my arguments? $\\$

E.g.

$$ \left(\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{matrix}\right) \begin{matrix} \to \\ \to \\ \to \\ \to \\ \end{matrix} \left(\begin{matrix} 5 \\ 6 \\ 7 \\ 8 \\ \end{matrix}\right) $$

But this mapping is not defined on e.g. {5,7}, which is a subset of $2^{Y}$.

$\endgroup$
5
  • 1
    $\begingroup$ It’s simply notation. By $f^{-1}(G)$ we mean the set of preimages $f^{-1}(G)$ for all $G$ in $\mathcal{G}$. $\endgroup$
    – user512346
    Jun 18, 2019 at 21:13
  • $\begingroup$ I did not get it. $\mathit f^{-1}$ is defined on $2^{Y}$? Do you mean that the domain of $\mathit f$ is $2^X$ and the codomain is $2^Y$? May be you meant to write $\mathit f^{-1}(\mathcal G)$, not $\mathit f^{-1}(G)$? In any case you provided no definition of notation which I suppose I do not inderstand. $\endgroup$ Jun 18, 2019 at 21:32
  • 1
    $\begingroup$ For $G \subset Y$, $f^{-1}(G) = \{x \in X : f(x) \in G\}$. $\endgroup$
    – snar
    Jun 18, 2019 at 21:35
  • $\begingroup$ Yes, I understand the definition of $\mathit f^{-1}(G)$. But the lemma is about $\mathit f^{-1}(\mathcal G)$, not about $\mathit f^{-1}(G)$. If $\mathcal G=\{\{5,7\},\{5\}\}$ in my example, what is $\mathit f^{-1}(\mathcal G)$? $\endgroup$ Jun 18, 2019 at 21:42
  • $\begingroup$ @snar here is the link of the lemma proofwiki.org/wiki/… $\endgroup$ Jun 18, 2019 at 22:07

1 Answer 1

2
$\begingroup$

OP agrees that $f^{-1}(G) = \{x \in X : f(x) \in G\}$ for $G \subset Y$, and the definition of $f^{-1}(\mathcal{G})$ was given in words in the comments as "the set of preimages $f^{-1}(G)$ for all $G \in \mathcal{G} $ ". Written verbatim, $$f^{-1}(\mathcal{G}) = \{f^{-1}(G) : G \in \mathcal{G}\} \subset \mathcal{P}(X).$$ The first step of the linked proof, $f^{-1}(\mathcal{G}) \subset f^{-1}(\sigma(\mathcal{G}))$ is spelled out as \begin{align*} f^{-1}(\mathcal{G}) &= \{f^{-1}(G) : G \in \mathcal{G}\} \\ &\subset \{f^{-1}(G) : G \in \sigma(\mathcal{G})\} \\ &= f^{-1}(\sigma(\mathcal{G})). \end{align*}

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$ Jun 18, 2019 at 22:19

You must log in to answer this question.

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