Image of an open set not containing $0$ under the map $x^2$ is open Let us consider the continuous function $\>\>\>f(x)=x^2$, on $\Bbb{R}$. Let $V$ be an open set in $\mathbb{R}$ such that $0 \notin V$. Then prove that $f(V)$ is open in $\mathbb{R}.$
So I tried to show that $\>\>\mathbb{R} - f(V)\>$ is closed. Let $\{x_n\}$ be a sequence in $\>\>\mathbb{R} - f(V)\>$ such that $x_n \to x.$ We need to show that $x \in $$\>\>\mathbb{R} - f(V)\>$. Now let $x \in f(V)$, then $$f(z)=x,\>\>\>\> z\in V.$$ This follows that $x>0.$ How can I proceed from here? Please help.
 A: I would do it as follows. Take $x\in V$. Now, take $a,b\in\mathbb R$ such that $x\in(a,b)\subset V$. Then either $(a,b)\subset(0,\infty)$ or $(a,b)\subset(-\infty,0)$. In the first case, $f\bigl((a,b)\bigr)=(a^2,b^2)$; otherwise, $f\bigl((a,b)\bigr)=(b^2,a^2)$. In any case, $f(V)$ is a neighborhood of $f(x)$. Since this occurs for each $x\in V$, $f(V)$ is an open set.
A: Let $f_\pm := f|_{\mathbb R_\pm}$ and $V_\pm := V\cap\mathbb R_\pm$ (both open). Then $f_\pm : \mathbb R_\pm\to\mathbb R_+$ has the continuous inverse $g_\pm = \pm\sqrt\cdot$. Hence
$$
f(V) = f_+(V_+)\cup f_-(V_-) = g_+^{-1}(V_+)\cup g_-^{-1}(V_-),
$$
so that $f(V)$ is indeed open.
A: Let $V$ open that does not contain zero.
Then $V$ is  a  union of  intervals $\bigcup_{c\in C}(a_c,b_c)$ for some index set $C$ where $-\infty \leq a_c<b_c \leq +\infty.$
So $f(V)=\bigcup_{c \in C}f(a_c,b_c)$
1.If some bounded interval  $(a_c,b_c)\subseteq (-\infty,0)$ then $f((a_c,b_c))=(b_c^2,a_c^2)$
2.If some bounded interval  $(a_c,b_c)\subseteq (0,+\infty)$ then $f((a_c,b_c))=(a_c^2,b_c^2)$
Also if some  interval is unbounded i.e of the form $(a_c,+\infty)$ or $(-\infty,b_c)$
Then
1.If $a_c<0$ then $f(a_c,+\infty)=f(a_c,0) \cup f(0,+\infty)=(0,+\infty)$
2.If $b_c<0$ $f(-\infty,b_c)=(b_c^2,+\infty)$
3.If $a_c>0$ then $f(a_c,+\infty)=(a_c^2,+\infty)$
4.If $b_c>0$ then $f(-\infty,b_c)=f(-\infty,0)\cup f(0,b_c)=(0,+\infty)$
So $f(V)$ is a union of open intervals thus open.
