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So by Urysohn's lemma: A topological space is normal (i.e., any two disjoint closed subsets can be separated by two disjoint open neighborhoods) iff there exists for any closed subsets $A,B$ a continuous function $f:X\rightarrow [0,1]$ such that $f_{|_A}=0$ and $f_{|_B}=1$.

Now, let $\mathbb R_l$ be the lower limit topology generated by the intervals $\{[a,b);\ a<b\}$. It is known that the space $\mathbb R_l\times \mathbb R_l$ is not normal. The proof of this fact is pretty much using the definition of normal spaces.

I was wondering if we can find two closed subsets $A,B$ of $\mathbb R_l\times \mathbb R_l$ such that there exists no continuous $f:\mathbb R_l\times \mathbb R_l\rightarrow [0,1]$ such that $f_{|_A}=0$ and $f_{|_B}=1$.

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You can find such $A$ and $B$ by a counting argument: let $D$ be the anti-diagonal of $\mathbb{R}_l \times \mathbb{R}_l$, so that $D = \{(x,-x): x \in \mathbb{R}_l \}$.

One can check that $D$ is closed and discrete as a subspace: e.g. $[x,x+1) \times [-x, -x+1) \cap D = \{(x, -x)\}$ so all singletons in $D$ are relatively open.

So for all subsets $A$ of $D$, $A$ and $D\setminus A$ are closed subsets of $\mathbb{R}_l \times \mathbb{R}_l $. So if the product were normal we'd have a continuous function $f_A: \mathbb{R}_l \times \mathbb{R}_l \to \mathbb{R}$ with $f_A[A] = \{0\}$ and $f_A[D\setminus A] = \{1\}$.

Note that the set $\{f_A: A \subseteq D\}$ has the same size as the powerset of $D$, i.e. $2^\mathfrak{c}$.

But $\mathbb{R}_l \times \mathbb{R}_l $ is separable, as $Q:=\mathbb{Q} \times \mathbb{Q}$ is dense, so all $f_A|Q$ should be different too (a function is uniquely determined by its values on a dense set) and there are only $\mathfrak{c}$ many real-valued functions with domain $Q$.

This contradiction shows we cannot find such $f_A$ for most subsets of $D$ in fact.

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