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By SIDNEY A. MORRIS

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Let T be an open equilateral triangle with base parallel to the X-axis and let y be any point in T . Then there exists an open rectangle R such that y ∈ R ⊆ T . Pictorially, this is again easy to see. Y .... ... . ...... ...................... ... . ......................... . . . . ............................................. ............................. ................................................... .. . . . . . . . ...................................... ............................

Thus x ∈ Bj ⊆ U , as required. Conversely, assume that for each U ∈ τ and each x ∈ U , there exists a B ∈ B with x ∈ B ⊆ U . We have to show that every open set is a union of members of B. So let V be any open set. Then for each x ∈ V , there is a Bx ∈ B such that x ∈ Bx ⊆ V . Clearly V = x∈V Bx. ) Thus V is a union of members of B. 3 Proposition. Let B be a basis for a topology τ on a set X. Then a subset U of X is open if and only if for each x ∈ U there exists a B ∈ B such that x ∈ B ⊆ U . Proof.

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