Urysohn's lemma in topology states that
- if X is a normal topological space and A and B are disjoint closed subsets of X, then there exists a continuous function from X into the unit interval [0, 1],
- f : X → [0, 1],
- such that f(a) = 0 for all a in A and f(b) = 1 for all b in B.
Note that in the statement above, we do not, and in general cannot, require that f(x) ≠ 0 and ≠ 1 for x outside of A and B. This is only possible in perfectly normal spaces.
Proof sketch
For every dyadic fraction r ∈ (0,1), we are going to construct an open subset U(r) of X such that:
- U(r) contains A and is disjoint from B for all r
- for r < s, the closure of U(r) is contained in U(s)
In order to construct the sets U(r), we actually do a little bit more: we construct sets U(r) and V(r) such that
- A ⊆ U(r) and B ⊆ V(r) for all r
- U(r) and V(r) are open and disjoint for all r
- for r < s, V(s) is contained in the complement of U(r) and the complement of V(r) is contained in U(s)
This construction proceeds by mathematical induction. Since X is normal, we can find two disjoint open sets U(1/2) and V(1/2) which contain A and B, respectively. Now assume that n≥1 and the sets U(a/2n) and V(a/2n) have already been constructed for a = 1,...,2n-1. Since X is normal, we can find two disjoint open sets which contain the complement of V(a/2n) and the complement of U((a+1)/2n), respectively. Call these two open sets U((2a+1)/2n+1) and V((2a+1)/2n+1), and verify the above three conditions.
The Mizar project has completely formalized and automatically checked a proof of Urysohn's lemma in the URYSOHN3 file.