Intersection and union under arbitrary function

Question

I have tried to grasp topology homeomorphism. I have heard that such transformations preserve intersections and unions. I have tried to prove it myself and it came out that actually any transformation preserves unions and intersections.

$$f: X \to Y $$ $$U, V \subset X $$ $ f(U \cup V)= f(${$a\in X \mid a \in U \vee a \in V $}$) = ${$f(a)\in Y \mid a \in U \vee a \in V $}$ = ${$f(a) \in Y \mid a \in U $}$ \cup ${$ f(a) \in Y \mid a \in V $}$ = f(U) \cup f(V) $

$ f(U \cap V)= f(${$a\in X \mid a \in U \wedge a \in V $}$) = ${$f(a)\in Y \mid a \in U \wedge a \in V $}$ = ${$f(a) \in Y \mid a \in U $}$ \cap ${$ f(a) \in Y \mid a \in V $}$ = f(U) \cap f(V) $

It feels off, so I am here to check if there is a mistake. Otherwise I would be glad to hear explanation from another point of view.

Answer 1

The simplest counterexample that I could imagine is simply, choose $U,V$ to be non-empty disjoint subsets of some set $X$ and define

$ f:X\to \{1\} $

Then $ f(U\cap V) = f(\emptyset) = \emptyset$

But

$f(U)\cap f(V) = \{1\}$

As for where your proof went wrong, I think it is your mistake is quite instructive, you make the assertion that

$\{f(a)\in Y| a\in V \land a\in U\} \\ = \{f(a)\in Y|a\in U\} \cap \{f(a)\in Y|a\in V\} $

But the problem is that in general,

$\{g(x)\in S|\alpha(x)\}$

Is not well defined, so you cannot blindly apply the definition of the intersection.

The axiom of restricted comprehension in $ZFC$ guarantees that the object

$\{x\in S|\alpha(x)\}$

Always forms a set, but when you "apply" a function to the first appearance of $x$ in the above expression, there is no longer any axiom that guarantees that this is a set, and for good reason, in general, we cannot deduce what $x$ is if we only know what $f(x)$ is, and so there is no way to verify $\alpha(x)$. When can we deduce what $x$ is from $f(x)$ ? Precisely when $f$ is injective, and not surprisingly, the intersection property you were trying to prove is true precisely for injective functions.

The union property is true for any function, but again I would advise caution, and to rather use the formal definition of image,

$f(X) = \{y\in Y|\exists_{x\in X}[f(x)=y]\}$

When you are unsure

Answer 2

Let me bring one example to help you:

Let $A=\left[ 0, \frac{3\pi}{4}\right]$ and $B=\left[ \frac{\pi}{2}, \frac{3\pi}{2}\right]$. If we take $f(x)=\sin x$, then $ f(A)\cap f(B) \not\subset f(A \cap B)$

What about all other variants?