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Date: 8-6-2021
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Definition: A topological space X is said to be a Hausdorff space if and only if it satisfies the following Hausdorff Axiom:
• if x and y are distinct points of X then there exist open sets U and V such that x ∈ U, y ∈ V and U ∩ V = ∅.
Lemma 1.1 All metric spaces are Hausdorff spaces.
Proof Let X be a metric space with distance function d, and let x and y be points of X, where x ≠y. Let ε =1/2d(x, y). Then the open balls BX(x, ε) and BX(y, ε) of radius ε centred on the points x and y are open sets (see Lemma 1.1). If BX(x, ε) ∩ BX(y, ε) were non-empty then there would exist z ∈ X satisfying d(x, z) < ε and d(z, y) < ε. But this is impossible, since it would then follow from the Triangle Inequality that d(x, y) < 2ε, contrary to the choice of ε. Thus x ∈ BX(x, ε), y ∈ BX(y, ε), BX(x, ε) ∩ BX(y, ε) = ∅.
This shows that the metric space X is a Hausdorff space.
We now give an example of a topological space which is not a Hausdorff space.
Example :The Zariski topology on the set R of real numbers is defined as follows: a subset U of R is open (with respect to the Zariski topology) if and only if either U = ∅ or else R U is finite. It is a straightforward exercise to verify that the topological space axioms are satisfied, so that the set R of real numbers is a topological space with respect to this Zariski topology. Now the intersection of any two non-empty open sets in this topology is always non-empty. (Indeed if U and V are non-empty open sets then U = R F1and V = R F2, where F1 and F2 are finite sets of real numbers. But then U ∩ V = R (F1 ∪ F2), which is non-empty, since F1 ∪ F2 is finite and R is infinite.) It follows immediately from this that R, with the Zariski topology, is not a Hausdorff space.
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دراسة يابانية لتقليل مخاطر أمراض المواليد منخفضي الوزن
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اكتشاف أكبر مرجان في العالم قبالة سواحل جزر سليمان
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اتحاد كليات الطب الملكية البريطانية يشيد بالمستوى العلمي لطلبة جامعة العميد وبيئتها التعليمية
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