3
$\begingroup$

$f:\mathbb R^2\rightarrow \mathbb R^2$ is any function and $G_1$ and $G_2$ are subsets of $\mathbb R^2.$ Then

  1. $f^{-1}(G_1\cup G_2)=f^{-1}(G_1)\cup f^{-1}(G_2).$

  2. $f^{-1}(G_1^c)=(f^{-1}(G_1))^c$

  3. $f(G_1\cap G_2)=f(G_1)\cap f(G_2).$

  4. $G_1$ is open and $G_2$ is closed then $G_1+G_2=\{x+y : x\in G_1 ,y\in G_2\}$ is neither closed nor open.

  5. $f(G_1\cup G_2)=f(G_1)\cup f(G_2)$

  6. $f^{-1}(G_1\cap G_2)=f^{-1}(G_1)\cap f^{-1}(G_2)$

By using the method of set inclusion and reverse inclusion $1,3,5,6$ are proved. Here is the proof for $3:$

$$x\in f(G_1\cap G_2)\\ \implies f^{-1}(x)\in G_1\cap G_2\\ \implies f^{-1}(x)\in G_1\ \text{and} \ f^{-1}(x)\in G_2\\ \implies x\in f(G_1)\ \text{and} \ x\in f(G_2)\\ \implies x\in f(G_1)\cap f(G_2)\\ \implies f(G_1\cap G_2)\subset f(G_1)\cap f(G_2) $$

Conversely $$y\in f(G_1)\cap f(G_2)\\ \implies y\in f(G_1)\ \text{and} \ y\in f(G_2)\\ \implies f^{-1}(y)\in G_1\ \text{and}\ f^{-1}(y)\in (G_2)\\ \implies f^{-1}(y)\in G_1\cap G_2\\ \implies y\in f(G_1\cap G_2)\\ \implies f(G_1)\cap f(G_2)\subset f(G_1\cap G_2)$$

Together they imply $$f(G_1\cap G_2)=f(G_1)\cap f(G_2).$$ Similar proofs hold for $1,5,6.$

Now the problem is that if I take $G_1=\mathbb Q$ and $G_2=\mathbb Q^c$ and $f(x)=0\ \forall x\in \mathbb R$ then we find a contradiction to $3.$ So what happened? What was the wrong step in the proof written above? And because of this now I cannot say which one of $1,3,5,6$ are correct or not.

Also please help prove or disprove $2$ and $4$.

Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

The most fundamental problem here, I believe, is you're treating $f^{-1}$ as if it were an actual inverse function, rather than a set; this is implied by your choice of $\in$ rather than $\subset$.

A function is often not invertible, but if $f:X\longrightarrow Y$we may still talk about $f^{-1}(A)$ for sets $A\subset Y$. This is defined as:

$$f^{-1}(A)=\{x\in X\,\,|\,\,f(x)\in A\}$$

and sometimes, for a single element $y\in Y$, there is some notational abuse in writing $f^{-1}(y)$ rather than $f^{-1}\big(\{y\}\big)$.

This problem is manifested in your 'Conversely' part, specifically when you say that (correcting the relators):

$$y \in f(G_1) \implies f^{-1}(y) \subset G_1$$

A simple counterexample is taking $y=1$, $f(x)=x^2$ and $G_1=\{1\}$. Clearly, $1=y \in f(G_1)=\{1\}$. However, $f^{-1}(y)=\{-1,1\}\not\subset G_1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .