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

Is there a simple method to prove that the square of the Sorgenfrey line is not normal?

The method in the book is a little complex.

Could someone help me?

share|cite|improve this question
2 gives an example of a pair of closed disjoint sets that don't have disjoint open supersets in the Sorgenfrey plane (i.e., the square of the sorgenfrey line) – Host-website-on-iPage Jul 14 '12 at 13:13
up vote 6 down vote accepted

I always use Jones' lemma. It's a handy tool to show non-normality of other spaces as well. You need some basic facts:

1) Suppose $X$ is normal. For every pair $A$, $B$ of closed disjoint non-empty subsets in $X$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 0$ for $x \in A$ and $f(x) = 1$ for $x \in B$. This is often called Urysohn's lemma.

2) If $f,g: X \rightarrow Y$ are continuous, and $Y$ is Hausdorff, and for some dense subset $D$ of $X$ we have $f(x) = g(x)$ for all $x \in D$, then $f(x) = g(x)$ for all $x \in X$. (Proof sketch: if not for some $x$, pull back disjoint open neighbourhoods of $f(x)$ and $g(x)$, both of these intersect $D$ and $f$ and $g$ cannot agree on those points.) This implies:

2') The function $R$ that maps a continuous function $f$ from $X$ to $Y$ to a continuous function $R(f)$ from $D$ to $Y$ by restricting $f$ to $D$, is 1-1.


Jones' Lemma: If $X$ is normal and $D$ is dense and infinite in $X$ and $C$ is closed and discrete (in the subspace topology) in $X$ then (as cardinal numbers) $2^{|C|} \le 2^{|D|}$.

Proof: for every non-trivial subset $A$ of $C$, $A$ and $C \setminus A$ are disjoint, closed in $X$ (both are closed in $C$, as $C$ is discrete, and closed subsets of a closed set are closed in the large set.), so by 1) there is a continuous function $f_A$ on $X$ that maps $A$ to $0$ and $C \setminus A$ to $1$.

Note that this defines a family of distinct continuous functions (if $A \neq B$ then we can find a point in $A\setminus B$ or $B \setminus A$ that shows that $f_A \neq f_B$) from $X$ to $[0,1]$. But from 2' we know that there is a 1-1 mapping from the set of all continuous functions from $X$ to $[0,1]$ to the set of all continuous functions from $D$ to $[0,1]$ and the latter set is bounded in size by $[0,1]^D = (2^{|N|})^{D} = 2^{|N||D|} = 2^{|D|}$, and the last step holds as $D$ is infinite.

As we have a family of size $2^{|C|}$ (all non-trivial, i.e. non-empty, non-$C$, subsets of $C$) we conclude that $2^{|C|} \le 2^{|D|}$, and this concludes the proof.


a) The Sorgenfrey plane: using the antidiagonal $C = \{(x, -x): x \in \mathbb{R} \}$ and $D = \mathbb{Q} \times \mathbb{Q}$ as dense subset. As $2^{|C|} = 2^\mathfrak{c} > \mathfrak{c} = 2^{|D|}$, Jones' lemma says that $X$ cannot be normal.

b) The Niemytzki plane (or Moore plane) (see e.g. here) is not normal, with a similar computation, using $C$ the $x$-axis and $D$ the rational points in the upper halfplane.

share|cite|improve this answer

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.