Show that $f(A ∩ B) ⊆ f(A) ∩ f(B)$, can the relation be improved to equality?

Question

I have the following question:

Let $A,B$ be two subsets of a set $X$, and let $f : X → Y$ be a function. Show that $f(A ∩ B) ⊆ f(A) ∩ f(B)$. Is it true that the $⊆$ relation can be improved to $=$?

While I know that in these type of questions I should start by going back to the formal definition of everything stated and work from there I am having trouble stating these formally.

Usually a function can be written as ${\displaystyle \{(x,f(x)):x\in X\}}$ but how do I alter this to show that $x \subseteq A ∩ B$

Does the following work: ${\displaystyle \{(x,f(x)):x\in (A ∩ B) \subseteq X\}}$

In case it does, here is attempt:

${\displaystyle \{(x,f(x)):x\in (A ∩ B) \subseteq X\}}$

is equivalent to: ${\displaystyle \{(x,f(x)):x\in A \subseteq X \wedge x\in B \subseteq X\}}$

which in turn is equivalent to: ${\displaystyle \{(x,f(x)):x\in A \subseteq X\}} \wedge {\displaystyle \{(x,f(x)):x\in B \subseteq X\}}$

which is equivalent to $f(A) ∩ f(B)$

The biggest flaw in my work is that I am working as if I am showing equality, can someone guide me or give hints?

Does this answer your question? Prove $f(S \cap T) \subseteq f(S) \cap f(T)$ — saru, May 29 '20 at 15:17
@devianceee this was helpful but my question addressed the notion of equality as well which is why I will keep it as someone already answered) — Sergio, May 29 '20 at 15:24
If you have any question also similar to this on preimages and images, you can refer to this giant list here. Although, I'd recommend that you try solving the questions to the fullest first. — saru, May 29 '20 at 15:29
The relation can be improved to equality if $f$ is injective. — Geoffrey Trang, May 29 '20 at 15:36

score 1 · Accepted Answer · answered May 29 '20 at 15:20

1

The first half of proving the containment is addressed by @devianceee's comment. Listed here (plz give his/her comment upvote if this was useful to you: Prove $f(S \cap T) \subseteq f(S) \cap f(T)$ )

The second half of the question asks about improving the containment to an equality.

I don't believe the containment can be improved to an equality.

Consider $f(x) = x^2$ On the set $A = \lbrace -1, 2 \rbrace$ and $B=\lbrace 1, 2 \rbrace$

We have that $A \cap B = \lbrace 2 \rbrace$

So $f(A \cap B) = \lbrace 4 \rbrace$

Now $f(A) = \lbrace 1,4 \rbrace$ and similarly $f(B) = \lbrace 1,4 \rbrace$

so $f(A) \cap f(B) =\lbrace 1,4 \rbrace$

Clearly $\lbrace 1,4 \rbrace \ne \lbrace 4 \rbrace$ so we do NOT have that

$ f(A) \cap f(B) = f(A \cap B)$

In general.

Interesting questions to ask:

What conditions on $f$ and the sets $A,B$ need to be added so that equality again becomes possible? How can we play around with those conditions. This can lead to an interesting rabbit hole I believe.

answered May 29 '20 at 15:20

Sidharth Ghoshal

16,771

I think equality would be in the case of the function being one to one, or $A=B$ or $A U B=\emptyset$ , am I right? – Sergio May 29 '20 at 15:25
one more question if I may ask is how I defined the set correct? – Sergio May 29 '20 at 15:26
Yes its reasonable to consider a function as a set of pairs. Formally speaking all functions are relations. And all relations are just sets of pairs, so a function naturally is a set of pairs as well. – Sidharth Ghoshal May 29 '20 at 15:29
thank you @forgeyedpeas – Sergio May 29 '20 at 15:31
Indeed, the problem with the $x^2$ map is that it is not injective. – Geoffrey Trang May 29 '20 at 15:36
@GeoffreyTrang but regardless if it is injective wouldn't equality hold if $=$ or $=∅$ – Sergio May 29 '20 at 16:03

Show that $f(A ∩ B) ⊆ f(A) ∩ f(B)$, can the relation be improved to equality?

1 Answers1

Interesting questions to ask: