Can you find a counterexample for this set theory statement?

[Romono]

How would you disprove if z ∈ (f(X) ∩ f(Y)) then z ∈ f(X ∩ Y)? (Where f: A -> B, if X, Y ⊆ A.) I'm just not sure how to approach this.

 

[Euge]

How would you disprove if z ∈ (f(X) ∩ f(Y)) then z ∈ f(X ∩ Y)? (Where f: A -> B, if X, Y ⊆ A.) I'm just now sure how to approach this.

Hi again Romono,

When facing questions like these, try starting with a function that has a finite domain and codomain. In this question for example, you can consider the function $f : \{1, 2\} \to \{1, 2\}$ given by $f(1) = 1$ and $f(2) = 1$. Let $X = \{1\}$ and $Y = \{2\}$. Then $X \cap Y = \emptyset$, so $f(X\cap Y) = \emptyset$. On the other hand, since $f(X) = \{1\}$ and $f(Y) = \{1\}$, $f(X) \cap f(Y) = \{1\}$. So we have $1 \in f(X) \cap f(Y)$, but $1\notin f(X \cap Y)$. Can you find another example?

 

[Nono713]

In general an easy way to come up with a counterexample here is to exploit the fact that the image of the empty set under any function $f$ is empty. Hence it suffices to find a function $f$ and two disjoint $X$ and $Y$ such that $f(X) \cap f(Y)$ is non-empty, and you are done. The constant function is probably the simplest example.​