Moore plane

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In mathematics, the Moore plane, also sometimes called Niemytzki plane (or Nemytskii plane, Nemytskii's tangent disk topology), is a topological space. It is a completely regular Hausdorff space (also called Tychonoff space) which is not normal. It is named after Robert Lee Moore and Viktor Vladimirovich Nemytskii.


Open neighborhood of the Niemytzki plane, tangent to the x-axis

If \Gamma is the upper half-plane \Gamma = \{(x,y)\in\R^2 | y \geq 0 \}, then a topology may be defined on \Gamma by taking a local basis \mathcal{B}(p,q) as follows:

  • Elements of the local basis at points (x,y) with y>0 are the open discs in the plane which are small enough to lie within \Gamma. Thus the subspace topology inherited by \Gamma\backslash \{(x,0) | x \in \R\} is the same as the subspace topology inherited from the standard topology of the Euclidean plane.
  • Elements of the local basis at points p = (x,0) are sets \{p\}\cup A where A is an open disc in the upper half-plane which is tangent to the x axis at p.

That is, the local basis is given by

\mathcal{B}(p,q) = \begin{cases} \{ U_{\epsilon}(p,q):= \{(x,y):  (x-p)^2+(y-q)^2 < \epsilon^2 \} \mid \epsilon > 0\}, & \mbox{if }  q > 0;  \\ \{ V_{\epsilon}(p):= \{(p,0)\} \cup \{(x,y):  (x-p)^2+(y-\epsilon)^2 < \epsilon^2 \} \mid \epsilon > 0\},  & \mbox{if } q = 0. \end{cases}


Proof that the Moore plane is not normal[edit]

The fact that this space M is not normal can be established by the following counting argument (which is very similar to the argument that the Sorgenfrey plane is not normal):

  1. On the one hand, the countable set S:=\{(p,q) \in \mathbb Q\times \mathbb Q: q>0\} of points with rational coordinates is dense in M; hence every continuous function f:M\to \mathbb R is determined by its restriction to S, so there can be at most |\mathbb R|^ {|S|} = 2^{\aleph_0} many continuous real-valued functions on M.
  2. On the other hand, the real line L:=\{(p,0): p\in \mathbb R\} is a closed discrete subspace of M with  2^{\aleph_0} many points. So there are 2^{2^{\aleph_0}} > 2^{\aleph_0} many continuous functions from L to \mathbb R. Not all these functions can be extended to continuous functions on M.
  3. Hence M is not normal, because by the Tietze extension theorem all continuous functions defined on a closed subspace of a normal space can be extended to a continuous function on the whole space.

In fact, if X is a separable topological space having an uncountable closed discrete subspace, X cannot be normal.

See also[edit]