Prove (if the statement is correct) $f^{-1} (\bigcap_{i \in J} B_{i} ) = \bigcap_{i \in J} f^{-1}(B_{i}).$

Question

Let $f : S \rightarrow T$ be a function. If $\{B_{i} | i \in J \}$ is a set of subsets of $T$, Show that $f^{-1} (\bigcap_{i \in J} B_{i} ) = \bigcap_{i \in J} f^{-1}(B_{i}).$

Sometimes I am given wrong statements to try to prove and then get stuck at a point that help me correct the erroneous statement.

I feel like this statement is incorrect, but I am unable to create an example to show this, could anyone help me in clarifying what conditions should my example satisfy and what it should not? so that I could create it.
Will the proof differ if the indexing set $J$ is countable or uncountable, if so how? (this question is not answered in the suggested question)

@KaviRamaMurthy how this verification differs between countable and uncountable sets? — Intuition, Jan 06 '20 at 07:51
Does this answer your question? What are the strategies I can use to prove $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$? — Emilia Dunfelt, Jan 06 '20 at 08:07
@EmiliaDunfelt No, the second question is not answered in the suggested question Also the first question is not answered clearly. — Intuition, Jan 06 '20 at 08:14

William Elliot · Answer 1 · 2020-01-07T06:54:43.173

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Let $C$ be a collection of sets.
Theorem: $f^{-1}$($ \cap$ $C$) = $ \cap$ { $f^{-1}(A) : A \in$ $C$ }.
Proof:
x in $f^{-1}$($\cap$ $C$) iff
$f(x) \in \cap$ $C$ iff
for all $A \in C$, $f(x) \in A$ iff
for all $A \in C$, $x \in f^{-1}(A)$ iff
$x \in \cap$ { f$^{-1}(A) : A \in C $}.

edited Jan 07 '20 at 06:54

answered Jan 06 '20 at 08:13

William Elliot

17,598

I think this does not answer all my questions – Intuition Jan 06 '20 at 08:23
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Things might clear up a bit more for the OP if an index set is used. – drhab Jan 06 '20 at 09:53
@Secretly. It does so where I used C for the index set. – William Elliot Jan 06 '20 at 11:46

score 1 · Accepted Answer · answered Jan 06 '20 at 10:26

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For any $x$ we have $$x\in f^{-1}(\cap_{i\in I}B_i)\iff$$ $$f(x)\in \cap_{i\in I}B_i\iff$$ $$ \forall i\in I\,(f(x)\in B_i)\iff$$ $$ \forall i\in I\,(x\in f^{-1}B_i) \iff$$ $$ x\in \cap_{i\in I}f^{-1}B_i.$$

answered Jan 06 '20 at 10:26

DanielWainfleet

57,985

Why did not we use induction in the proof? – Intuition Jan 06 '20 at 10:36
@Secretly . Because $I$ can be any set, finite or infinite. – DanielWainfleet Jan 06 '20 at 16:37
do you mean countable or uncountable? I think induction can be done for finita as well as infinite sets – Intuition Jan 07 '20 at 03:15
ANY set......... – DanielWainfleet Jan 08 '20 at 07:22

Prove (if the statement is correct) $f^{-1} (\bigcap_{i \in J} B_{i} ) = \bigcap_{i \in J} f^{-1}(B_{i}).$

2 Answers2