Let $ f : \mathbb{R}^2 \rightarrow \mathbb{R} : (x,y) \mapsto x + y^2 $

Does $f$ map open sets to open sets with respect to the standard topology on $\mathbb{R}$?

I welcome any answers to this problem, below is my attempt.

Any open set in $\mathbb{R}^2$ can be written as a union of open balls in $\mathbb{R}^2$, this is because the collection of open balls form a topological basis. Also the image of a union of sets is the union of the images of each set and a union of open sets is open. Thus the problem reduces to:

Is $\{x + y^2 | (x,y) \in B_r(a,b) \}$ an open set?

Special case:

$\{x + y^2 | (x,y) \in B_1 (0,0) \} = (-1,1.25)$

This was obtained by noting

$\sup \{x + y^2 | (x,y) \in B_1 (0,0) \} = \\ \max \{x + y^2 | x^2 + y^2 =1 \} = \max \{\cos (\theta) + \sin^2(\theta) | \theta \in [0,2\pi] \}$

And the last expression can be solved by Weierstrass' Extreme Value theorem.

  • 2
    $\begingroup$ The notation $(x,y) \in B_r (x,y)$ is not very fortunate. It's like writing $\int_0^t f(t)\, dt$. $\endgroup$
    – PhoemueX
    Sep 22, 2015 at 7:03

2 Answers 2


Let $\Omega\subset{\mathbb R}^2$ be an open set, and consider a point $c_0\in f(\Omega)$. There is a point $(x_0,y_0)\in\Omega$ with $f(x_0,y_0)=c_0$. As $\Omega$ is open the is an $h>$ such that the segment $\sigma_h$ with endpoints $(x_0-h,y_0)$ and $(x_0+h,y_0)$ is contained in $\Omega$. It follows that $$[c_0-h,c_0+h]=f(\sigma_h)\subset f(\Omega)\ .$$ Since $c_0\in f(\Omega)$ was arbitrary, this proves that $f(\Omega)$ is open.

Note that we did use the exact definition of $f$; continuity would not be sufficient.


Let $(u,v) \in U$, with $U \subset \Bbb{R}^2$ open. We need to show that there is $\epsilon >0$ with $B_\epsilon (f(u,v))\subset f(U)$.

For this, simply note that there is $\epsilon >0$ with $B_\epsilon (u) \times\{ v \}\subset U$.

Together with the special form of $f$, this proves the claim (why?).


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