Is there a continuous bijection from open unit interval to open unit disc?

by Brahadeesh   Last Updated June 20, 2018 13:20 PM

I came across a problem asking the reverse of the titular question:

Prove that there is no continuous bijection from $B_1(\mathbb{0}) = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 < 1\}$ to $(0,1)$.

This I did as follows:

Suppose for the sake of contradiction that there is such a map $f$. Consider the map $g : B_1(\mathbb{0}) \setminus \{ \mathbb{0} \} \to (0,1) \setminus \{ f(\mathbb{0}) \}$ given by $g(\mathbb{x}) = f(\mathbb{x})$. Then $g$ is also a continuous bijection. The continuous image of a connected set must be connected, and the domain is connected, so the image must be contained in either $(0,f(\mathbb{0}))$ or $(f(\mathbb{0}),1)$, which contradicts surjectivity. Hence, proved.


I then decided to try and prove the claim in the other direction:

Q. Is there a continuous bijection $f : (0,1) \to B_1(\mathbb{0})$?

But, the same idea failed to work.

After removing a point from the domain and its image from the codomain, I have two disjoint intervals on the one hand and a connected space on the other. I tried taking a path in the codomain connecting two points, each lying in the image of the two components of the domain. But since the inverse of $f$ is not assumed to be continuous, I couldn't proceed further.

I'm not sure what to do next. Can someone help me prove or disprove the claim? Thanks!



Answers 1


Note that $(0,1)$ is the countable union of the compact sets $\left[\frac1n,1-\frac1n\right]$ ($n>2$). It turns out that the interior of $C_n=f\left(\left[\frac1n,1-\frac1n\right]\right)$ is empty. Suppose otherwise. Since $\left[\frac1n,1-\frac1n\right]$ is compact and $B_1(0)$ is Hausdorff, $f$ induces a homeomorphism from $\left[\frac1n,1-\frac1n\right]$ onto $C_n$. Therefore, the restriction of $f^{-1}$ to an open ball contained in $C_n$ is a homeomorphism from that open ball in $B_1(0)$ onto a subset of $(0,1)$. That cannot be by the connectedness argument that you mentioned.

So, $f\bigl((0,1)\bigr)$ is a countable union of closed sets with empty interior. Now, the Baire Category Theorem tells us that $f\bigl((0,1)\bigr)$ cannot be $B_1(0)$.

José Carlos Santos
José Carlos Santos
June 20, 2018 12:40 PM

Related Questions


Use continuity to prove inequality in $\textbf{R}^n$

Updated February 28, 2017 04:20 AM

Continuity of functions inside a open ball

Updated July 25, 2017 03:20 AM

Showing a function is continuous(real analysis)

Updated October 03, 2017 20:20 PM

Proving a function (on metric space) is continuous

Updated October 04, 2017 05:20 AM