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Urysohn's LemmaUrysohn'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 1,
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- f : X → 1,
- such that f(a) = 0 for all a in A and f(b) = 1 for all b in B.
The lemma, sometimes called "the first non-trivial fact of point set topology", is often used to construct continuous functions with various properties; it is widely applicable since all metric spaces and all compact Hausdorff spaces are normal. The lemma is generalized by (and usually used in the proof of) the Tietze extension theorem. 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. The lemma is named after Paul Samuilovich Urysohn, 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)
Once we have these sets, we define f(x) = inf { r : x ∈ U(r) } for every x ∈ X. Using the fact that the dyadic rationals are dense, it is then not too hard to show that f is continuous and has the property f(A) ⊆ {0} and f(B) ⊆ {1}. 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)
Since the complement of V(r) is closed and contains U(r), the latter condition then implies condition (2) from above. 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.
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