- #1

zacharyh

- 7

- 0

**Homework Statement**

__General Case:__

Let [tex]f:A \rightarrow B[/tex] be a function. Show that [tex]f( \bigcap_{\alpha\in\Lambda} T_{\alpha}) = \bigcap_{\alpha\in\Lambda} f(T_{\alpha})[/tex] for all choices of [tex]\{T_{\alpha}\}_{\alpha\in\Lambda[/tex] if and only if [tex]f[/tex] is one-to-one.

__Simpler Case:__

Let [tex]f:A \rightarrow B[/tex] be a function, and [tex]X, Y[/tex] be subsets of [tex]A[/tex].

Show that [tex]f(X \cap Y) = f(X) \cap f(Y)[/tex] if and only if [tex]f[/tex] is one-to-one.

**The attempt at a solution**

I am trying to understand the simpler case first before attempting a solution for the general case.

(<=)

Assume [tex]f[/tex] is one-to-one. Let [tex]b \in f(X) \cap f(Y)[/tex].

Then, [tex]b \in f(X), b \in f(Y)[/tex]. Then, [tex]\exists a (a \in X, a \in Y, f(a) = b)[/tex].

This implies that [tex]a \in X \cap Y[/tex]. Therefore, [tex]f(X \cap Y) = f(X) \cap f(Y)[/tex].

(=>)(Our hint here is to use contraposition -- assuming the following to show f is

**not**one-to-one.).

Assume [tex]f(X \cap Y) \neq f(X) \cap f(Y).[/tex] Let [tex]u \in f(X) \cap f(Y)[/tex].

Then, [tex]u \in f(X), u \in f(Y)[/tex]. This implies [tex]\exists a,b (a \in X, b \in Y, f(a) = f(b) = u)[/tex].

If I show that [tex] a,b \notin (X \cap Y) [/tex], does that mean [tex] a \neq b [/tex]?

If so, then, [tex] f(a) = f(b) [/tex], so [tex]f[/tex] cannot be one-to-one.

Any thoughts?