# $f:\mathbb R^2 \rightarrow \mathbb R^2$ and $G_1$ and $G_2$ are subsets of $\mathbb R^2$

$$f:\mathbb R^2\rightarrow \mathbb R^2$$ 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.

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, $f^{-1}(y)=\{-1,1\}\not\subset G_1$.