# Prove that a continuous function from $B(0,1) \subset \mathbb R^2$ to $\mathbb R$ can not be one-to-one.

Prove that a continuous function from $B(0,1) \subset \mathbb R^2$ to $\mathbb R$ can not be one-to-one where $B(x,\epsilon)=\{y \in \mathbb R^2: d(x,y)<\epsilon\}$

I want to prove use the fact that connected sets are mapped to connected sets. However, I am unsure of how to apply that because both of these sets are connected.

• HINT: Suppose $0$ is in the image and has exactly one preimage. Now can you think of something to do with those respective points so that continuity might tell you something? – Ted Shifrin Oct 27 '16 at 21:17
• Intuitively, if we have a function $f:E\to G$ with $\dim E > dim G$ You always loose some information when you apply $f$, which is the same as saying $f$ is non reversible or $f$ is surjective. – Furrane Oct 27 '16 at 21:18
• @Furrane I'm not quite sure this intuition works. Wouldn't that say that the case $f\colon \mathbb{R}^3\to\mathbb{R}^2$ is similar? – Clement C. Oct 27 '16 at 21:21
• Possible duplicate of Is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$ – Clement C. Oct 27 '16 at 21:22
• @KajHansen has deleted his answer, but I want to encourage him to undelete it because a slight modification works: Just consider $\phi$ restricted to $\overline {B(0,1/2)}.$ Because this set is compact, $\phi$ will be a homeomorphism onto $\phi (\overline {B(0,1/2)}).$ – zhw. Oct 27 '16 at 21:49

Suppose there were such a mapping $f.$ For each $y\in (0,1),$ consider $E_y = \{(x,y): 0< x < \sqrt {1-y^2}\}.$ Each $E_y$ is connected and none of them is a single point. Because $f$ is continuous and injective, each $f(E_y)$ is connected and not a single point. Hence each $f(E_y)$ is an interval of positive length. The collection $\{f(E_y): y \in (0,1)\}$ is thus an uncountable pairwise disjoint collection of intervals of positive length. That is a contradiction, because each $f(E_y)$ contains a rational, and there are only countably many of these.