Given a metric, find continuous map that maps to $\mathbb{R}^2$

Question

I wanted to know how to approach the question of:

Given a metric on $\mathbb{R}^2$ does a continuous map $f:\mathbb{R}^2 \to \mathbb{R}^2 $ with the following property:

$$f([0,1]\times[0,1]) = \mathbb{R}^2 $$

exist ?

I currently am a little clueless on how to think about it. I would guess we can work with the map of $f^{-1}$ on our map. But again not really sure. A little pointer in the right direction might do the trick though.

Any help is greatly appreciated.

To be clear, are you asking about the Euclidean metric or are you allowed to choose the metric as well? You only say "a metric". — paul blart math cop, Jul 14 '20 at 21:28

alphaomega · Accepted Answer · 2020-07-14T21:39:00.210

2

No such a function cannot exist. Continuous image of a compact set is a compact set, but $\mathbb{R}^2$ is not compact.

edited Jul 14 '20 at 21:39

answered Jul 14 '20 at 21:25

alphaomega

2,216

This answer is correct for the Euclidean topology, but as pointed out in the comments OP is allowing any metric. – paul blart math cop Jul 14 '20 at 21:52
yes that is correct. – alphaomega Jul 14 '20 at 21:58
and you already pointed out the example with the discrete metric. nice – alphaomega Jul 14 '20 at 21:58
@Mark Lauer This asnwer is correct. The question asks if $f$ exist for every metric and here is a metric for which there is no $f$. – Kavi Rama Murthy Jul 14 '20 at 23:24

score 2 · Answer 2 · answered Jul 14 '20 at 21:52

I'd suggest the discrete one: $d(x, y) = 1$ if $x \neq y$ and $d(x, y) = 0$ if $x = y$. The discrete metric induces the discrete topology, i.e. every subset is open. To see this, observe that the ball of radius $1/2$ centered at a point $x$ is precisely the singleton $\{x\}$. Every function out of a discrete space is continuous, since every subset is open. Thus, you need only find a function with $f[[0, 1] \times [0, 1]] = \mathbb R^2$. I'd rather not write one down explicitly, but however you have proven that these sets have the same cardinality will produce such a function. If you're unfamiliar with why these sets have the same cardinality, here are some relevant links:

bijection between open and closed intervals

bijection between open interval and \mathbb R

You can combine these to get a bijection $f: [0, 1] \times [0, 1] \longrightarrow \mathbb R^2$ which is then continuous in the discrete topology.

The way I read the question OP is asking if such a function exists for every metric. You cannot choose your metric. — Kavi Rama Murthy, Jul 14 '20 at 23:19

Given a metric, find continuous map that maps to $\mathbb{R}^2$

2 Answers2