Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this problems

  1. Proof that the ball $B_1{(0,0)}$ can be embedded in Moore plane (Niemytzki plane)
  2. Proof that $({\mathbb R}^3; \textrm{usual topology})$ can be embedded in Moore plane (Niemytzki plane)
  3. Proof that $({\mathbb R}; \textrm{usual topology})$ can be embedded in Moore plane (Niemytzki plane)
    For the second I dont know if the function $h(x)=\left(x,\arctan(x)+ \frac{\pi}{2}\right)$ is right.


share|cite|improve this question

The first part should be easy via $(x,y)\mapsto (x,y+1)$, the third via $x\mapsto (x,1)$.

However, I doubt that the second statement is true: Assume $f\colon \mathbb R^3\to \Gamma$ is an embedding. If there is a point $(x,y)\in\mathbb R^3$ such that $f(p)=(u,v)$ has $v>0$, then there is an open 3D ball around $(x,y)$ that is embedded into an open disc in $\mathbb R^2$. This is not possible (though I'd need homology to show that). Therefore $f(\mathbb R^3)\subseteq \{(u,v)\in\Gamma\mid v=0\}$, but the latter set is discrete.

Edit: after the second part is changed from $\mathbb R^3$ to $\mathbb R^2$, the statement is clear: $\mathbb R$ is homeomorphic to an open interval, e.g. $(-1,1)$, hence $\mathbb R^2$ to $(0,1)^2$, which can be placed inside $\Gamma$.

share|cite|improve this answer
My fault it is ${\mathbb R}^2$ – vic Oct 26 '12 at 17:11
Then please edit the question. – Hagen von Eitzen Oct 26 '12 at 18:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.