Moore plane / Niemytzki plane and Subspace Topology In the definition of the Moore plane X=L1∪L2, where L1 is the line y=0 and L2=X∖L1 , I have a problem. How  can i show using Subspace Topology that L1 is discreet?
 A: The answer to this question depends on how you defined the topology.
However, if $p = (x,0) \in L_1$, then a fundamental system of open neighborhoods for $p$ are balls centered at $(x,\epsilon)$ with radius $\epsilon$ (union the tiny point $p$).
When you intersect this open neighborhood with $L_1$, you obtain the singleton $\{p\}$!
A: A basic neighbourhood of a point $(x,y)$ where $y > 0$ is a classical (standard metric) ball of the plane $B((x, y), r)$, where $r \le |y|$. Note that all these balls miss $L_1$, so $X \setminus L_1$ is open, and $L_1$ is closed.
A basic neighbourhood of a point of the form $(x,0)$ is of the form 
$$N_r((x,0)) = \{(x,0)\} \cup B((x,r), r),r > 0 \text{ and note : } N_r((x,0) \cap L_1 = \{(x,0)\}$$  
which makes it clear that $\{(x,0)\}$ is open in $L_1$ by the definition of the subspace topology. A subspace is discrete iff each of its points is an isolated point in the subspace topology. So $L_1$ is closed and discrete, while it's clear that both the balls $B((x, y),r)$ and $N_r((x,0))$ intersect $\{(x,y): x,y \in \mathbb{Q}, y > 0\}$, which is thus a countable dense set.
So Jones' lemma will imply that $X$ is not normal.
