Show that $f(U) = \big\{ f(x_1,x_2) : (x_1,x_2) \in U \big\}$ is an open set of $\mathbb{R}$ with the standard metric.

Let $f:\mathbb{R}^2 \to \mathbb{R}$ such that $f(x_1,x_2)=x_2$. I am required to show that $$f(U)= \big\{ f(x_1,x_2) : (x_1,x_2) \in U \big\}$$ is an open set of $\mathbb{R}$ with the standard metric. Given that U is an open subset of $\mathbb{R}^2$. The metric on $\mathbb{R}^2$ is given by $$d((x_1,x_2),(y_1,y_2)):= \max \big\{\vert x_1-y_1\vert,\vert x_2-y_2\vert \big\}.$$ Can anyone please give me any hints with how to proceed with this question?

• How is $U$ given? – Hetebrij Apr 7 '16 at 17:29
• Is $U$ an open set in$\mathbb{R}^2$. ? – seeker Apr 7 '16 at 17:29
• The metric on $\mathbb{R}^2$ is not well-defined. – sqtrat Apr 7 '16 at 17:30
• Note that $f((a,b) \times (c,d)) = (c,d)$. Every point $x$ in $U$ is contained in some rectangle $(a,b) \times (c,d) \subset U$. – copper.hat Apr 7 '16 at 17:33
• @seeker Yes , sorry – anon Apr 7 '16 at 17:42

1 Answer

Suppose that $U$ is open in $\mathbb{R}^2$. If $f(U) = \emptyset$, we are done. So assume that $f(U)\not=\emptyset$. If $x_2 \in f(U)$, then there is an $(x_1,x_2)$ in $U$ for some $x_2$. Since $U$ is open there is an $\varepsilon >0$ such that $V:= (x_1-\varepsilon, x_1+\varepsilon)\times (x_2-\varepsilon, x_2+\varepsilon) =B_d((x_1,x_2),\varepsilon)\subset U$. Then $$(x_2-\varepsilon,x_2+\varepsilon) = f(V)\subset U$$ which is clearly an open subset of $\mathbb{R}$.

• Thanks for your reply. Can we just state that $f(V)=(x_2 - \epsilon, x_2 + \epsilon)$ ? – anon Apr 7 '16 at 17:51
• Yes, because $f$ is the projection map onto the second factor and the balls in the space under the metric $d$ are open squares. – sqtrat Apr 7 '16 at 17:54