Counter example for $f( \cap_{E \in \epsilon} E) \supseteq \cap_{E \in \epsilon} f(E)$

Question

I'm having difficulty wrapping my head around the idea of a collection of subsets. I'm trying to think of a counter example for $f( \cap_{E \in \epsilon} E) \supseteq \cap_{E \in \epsilon} f(E)$ where $\epsilon$ is a collection of subsets in X and $E \subset X$.

Answer 1

Let's use the function $f: \Bbb R \to \Bbb R$ defined by $f(x) = x^2$ and choose the sets $E_1 = (-\infty,0]$ and $E_2 = [0,\infty)$ with $\mathcal E = \{E_1,E_2\}$. Then

$$f\left(\bigcap_{E \in \mathcal E} E\right) = f(E_1 \cap E_2) = f\bigl((-\infty,0] \cap [0,\infty)\bigr)= f(\{0\}) = \{0\}$$

but

$$\bigcap_{E \in \mathcal E} f(E) = f(E_1) \cap f(E_2) = f\bigl((-\infty,0]\bigr) \cap f\bigl([0,\infty)\bigr) = [0,\infty) \cap [0,\infty) = [0,\infty).$$

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Note that if you used $(-\infty,0)$ and $(0,\infty)$, then your intersection for the left would actually be empty!

Also, you can find counterexamples for any function that is not one-to-one since if $f(x_1) = f(x_2)$, but $x_1 \ne x_2$, then $E_1 = \{x_1\}$ and $E_2 = \{x_2\}$. Thus $f(E_1 \cap E_2) = f(\varnothing) = \varnothing$, but $f(E_1) \cap f(E_2) = f(x_1) = f(x_2) \not\subseteq \varnothing$