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hw4sol-andy - 147 HW4 Solutions 17.15 Show the T1 axiom is...

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147 HW4 Solutions 17.15 Show the T 1 axiom is equivalent to the condition that for each pair of points of X , each has a neighborhood not containing the other. T 1 axiom: All finite point sets are closed. Solution: Assume the T 1 axiom. Then for any pair of poinits x, y X , X - { y } is a neighborhood of x not containing y . Assume that for each pair of points of X , each has a neighborhood not containing the other. Let x X . The for all y 6 = x in X . There exists a neighborhood U y of y not containing x . Thus { x } = X - S y 6 = x U y is a closed set. As finite intersections of closed sets are closed, we have the T 1 axiom. 17.20 Find the boundary and the interior of each of the following subsets of R 2 : (a) A = { x × y | y = 0 } Solution: This is the x -axis. As it contains no basis elements ( a, b ) × ( c, d ), its interior is empty and (from 17.19a), as it’s closed, it is its own boundary. (b) B = { x × y | x > 0 and y 6 = 0 } Solution: This is the union of the first and fourth open quadrants. So Int B = B and the y -axis and positive x -axis.
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