By Louis Auslander

Lawsuits of the yankee Mathematical Society

Vol. sixteen, No. 6 (Dec., 1965), pp. 1230-1236

Published by way of: American Mathematical Society

DOI: 10.2307/2035904

Stable URL: http://www.jstor.org/stable/2035904

Page count number: 7

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**Additional resources for An Account of the Theory of Crystallographic Groups**

**Example text**

We now collect some properties of limits. 18 (Uniqueness of limits in metric spaces). Let (X, d) be a metric space. If xk → p and xk → q as k → ∞, then p = q. Proof. Let ε > 0. Since xk → p and xk → q as k → ∞, it follows that there are some numbers N1 , N2 such that d(xk , p) < ε for all k > N1 and such that d(xk , q) < ε for all k > N2 . Hence by the triangle inequality for all k > max{N1 , N2 } we have d(p, q) ≤ d(p, xk ) + d(xk , q) < 2ε. Since this conclusion is true for any ε > 0, it follows that d(p, q) = 0 and hence p = q.

Third, let A = {Aj : j ∈ J} ⊂ τi . Now A Aj =i(Aj ) = i(Aj ) i(Aj )⊂i( ⊂ A) i A ⊂ A. j∈J Thus A = i( A) ∈ τi . Next, intτi (A) = {U ∈ τi | U ⊂ A} = {i(B) | i(B) ⊂ A, B ⊂ X} . Here we see that i(A) ⊂ intτi (A). Moreover, if i(A) ⊂ i(B) ⊂ A then i(A) = i(i(A)) ⊂ i(i(B)) = i(B) ⊂ i(A). Hence i(A) = intτi (A). 14. Above we have seen how topologies and closure operators (or interior operators) on a set are in bijective correspondence. 8. 15. For each j ∈ J, let τj be a topology on X. Prove that τ = τj j∈J is a topology.

On the other hand, it may well be that a countably inﬁnite intersection of open sets is not open. In a metric space (X, d), ∞ B1/k (x) = {x}. k=1 Now {x} ∈ τd if and only if {x} = Br (x) for some r > 0. 14 (Properties of closed sets). Let X be a topological space. We have the following properties of closed sets in X: (C1) (C2) (C3) ∅ and X are closed in X. The intersection of any collection of closed subsets of X is closed. The union of a ﬁnite collection of closed subsets of X is closed. Proof.

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