If $f(\bigcap_{\alpha\in{I}}A_\alpha)=\bigcap_{\alpha\in{I}}f(A_\alpha)$ then $f$ is an injection?

Question

I know that if $f$ is an injection then $f(\bigcap_{\alpha\in{I}}A_\alpha)=\bigcap_{\alpha\in{I}}f(A_\alpha)$ but, the inverse is true?. I mean If $f(\bigcap_{\alpha\in{I}}A_\alpha)=\bigcap_{\alpha\in{I}}f(A_\alpha)$ then $f$ is an injection

I would appreciate if you explain your answer please.

Thanks for your time.

@alejopelaez: then for two disjoint sets the claimed property does not hold — SBF, Apr 02 '13 at 14:52

score 4 · Accepted Answer · answered Apr 02 '13 at 14:54

4

Suppose $f(x_1)=f(x_2)$ for some $x_1\neq x_2$. Then $\{x_1\}\cap\{x_2\}=\emptyset$ so $f(\{x_1\}\cap\{x_2\})=\emptyset$. But $\{f(x_1)\}\cap\{f(x_2)\}\neq\emptyset$.

answered Apr 02 '13 at 14:54

Hui Yu

15,029

thanks @Hui Yu, I forgot to say that $(A_\alpha)_{\alpha\in{I}}$ is a family of subsets of $X$ and $f: X\to Y$. Therefore f is an inyection in $X$? – Fernando Valle Apr 02 '13 at 15:05
forgive it, I understand now. Thanks. – Fernando Valle Apr 02 '13 at 15:16
@FernandoValle That's fine. Welcome to the site. If you are satisfied, you can probably check the tick and accept my answer. – Hui Yu Apr 02 '13 at 16:25

If $f(\bigcap_{\alpha\in{I}}A_\alpha)=\bigcap_{\alpha\in{I}}f(A_\alpha)$ then $f$ is an injection?

1 Answers1