# Download Topology Course lecture notes by A. McCluskey, B. McMaster PDF

By A. McCluskey, B. McMaster

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Hence A ⊆ G. Similarly B ⊆ H. It’s true that T4 ⇒ T3 1 but not very obvious. e. G1 ⊇ (closed) X \ H ⊇ G 1 ). 1 (Urysohn’s Lemma) Let F1 , F2 be disjoint non-empty closed subsets of a T4 space; then there exists a continuous function f : X → [0, 1] such that f (F1 ) = {0}, f (F2 ) = {1}. Proof Given disjoint closed F1 and F2 , choose disjoint open G0 and H0 so that F1 ⊆ G0 , F2 ⊆ H0 . Define G1 = X\F2 (open). Since G0 ⊆ (closed) X\H0 ⊆ X \ F2 = G1 , we have G¯0 ⊆ G1 . By the previous remark, we can now construct: (i) G 1 ∈ T : G¯0 ⊆ G 1 , G¯1 ⊆ G1 .

If {Xi : i ∈ I} is any family of sets, then their product is {x : I → ∪i∈I Xi for which x(i) ∈ Xi ∀i ∈ I} except that we normally write xi rather than x(i). Then a typical element of X = Xi will look like: (xi )i∈I or just (xi ). We will still call xi the ith coordinate of (xi )i∈I . 2 Topologizing the Product Of the many possible topologies that could be imposed on X = Xi , we describe the most useful. 1. e. πi ((xi )i∈I ) = xi . e. πi−1 (Gi ) where i ∈ I, Gi = ∅, Gi ∈ Ti . (Gij ). ) 37 We use these open cylinders and boxes to generate a topology with just enough open sets to guarantee that projection maps will be continuous.

Conversely, if f is not continuous, there exists A ⊆ X such ¯ \ f (A) so p = f (x), ¯ ⊆ f (A). Thus, there exists p ∈ f (A) that f (A) ¯ So there exists a sequence (xn ) in A with xn → x. some x ∈ A. 34 Yet, if f (xn ) → f (x)(= p), p would be the limit of a sequence in f (A) so that p ∈ f (A) —contradiction! Thus f fails to preserve convergence of this sequence. 35 Chapter 4 Product Spaces A common task in topology is to construct new topological spaces from other spaces. One way of doing this is by taking products.