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Let $(\Bbb R,\tau)$ be the Sorgenfrey line. Why $\Bbb R^2$ is not normal?

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marked as duplicate by vonbrand, Davide Giraudo, arjafi, Amzoti, GEdgar Apr 14 '13 at 13:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

To be clear, by $\mathbb{R}^2$ you're referring to the set $\mathbb{R}^2$ with the product topology inherited from the product of two copies of the Sorgenfrey line? – Dan Rust Apr 14 '13 at 12:57
up vote 5 down vote accepted

$\mathbb{R}^2$ in this topology is separable (as witnessed by $\mathbb{Q}^2$) and has a closed and discrete (in itself, so discrete as a subspace) subspace $D = \{ (-x, x) : x \in \mathbb{R} \}$, where discreteness is witnessed by the fact that $[-x, -x+1) \times [x, x+1)$ intersects $D$ in $(-x,x)$ only.

Now all for all non-empty subsets $A$ of $D$ we know that $A$ and $D \setminus A$ are closed and disjoint in $D$ and thus also in $\mathbb{R}^2$, and so if the square were normal we would have a continuous function $f_A : \mathbb{R}^2 \rightarrow [0,1]$ that sends $A$ to $0$ and $D \setminus A$ to $1$, for all subsets $A$ of $D$. In particular, we have at least $2^{|\mathbb{R}|}$ many distinct continuous functions on $\mathbb{R}^2$ with values in $[0,1]$.

But if $X$ is a separable space with countable dense set $E$, then any two functions from $X$ into $[0,1]$ that coincide on $E$, coincide on the whole of $X$. This is a standard fact, and we only need Hausdorffness in the codomain. But this just says that the function that sends a function $f : X \rightarrow [0,1]$ to its restriction $f|E$ is an injection. But apply this to the separable space $\mathbb{R}^2$: we get that there are at most as many continuous functions into $[0,1]$ as there are continuous functions from $\mathbb{Q}^2$ into $[0,1]$, by this injection, and this set has cardinality (at most, in fact equal) $[0,1]^{\mathbb{Q}^2} = 2^{|\mathbb{N}|} = |\mathbb{R}|$.

So in short, if the square of the Sorgenfrey were normal, we'd need at least as many continuous functions on it as there are subsets of $\mathbb{R}$, but there are only $\mathbb{R}$ many by separability.

This argument is generalized in Jones' lemma: if $X$ is a normal space with a dense set $D$, then the size of a closed and discrete subspace $C$ obeys $|C| < 2^{|D|}$.

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Let $\Delta=\{(x,-x)|x\in\mathbb{R}\}$. For an accessible example of a pair of closed subsets that can't be seperated by open neighbourhoods, use the closed subsets $A=\{(x,-x)|x\in\mathbb{Q}\}$ and $\Delta\setminus A$.

Showing that $A$ and $\Delta\setminus A$ are closed does require some fiddling with the basis elements of the topology but is rather easy when you recall that the basis elements are given by $B_{a,b}=\{(x,y)|0\leq x<a,0\leq y<b\}$ and the translates of such 'half-open' rectangles. It is then easy to find an infinite union of such rectangles which give the usual 'fully-open' rectangles and then building the complement of $A$ with the union of such rectangles is no problem.

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