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Hausdorff and Non-Hausdorff Spaces

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Let E be a subset of R n . show that the derived set of E, Eâ€™, is closed Solution Suppose that X is a space such that all singleton subsets of X are closed. (this includes all metric spaces; this condition is known by the name T_1). Then for all subset E of X, E is closed, where A\' is the derived set of E. Recall that E\' is the set of all x such that every open neighbourhood of x (every open ball, if you are working metrically) intersects E in a point other than x. Proof: Suppose that x is NOT in E\'. This means that there is an open neighbourhood U of x, such that U /\\ E is either {x} or empty. Now, every point y of U is also NOT in E\': for x this is clear, for all other y we use U\\{x}, which is open as X has closed singletons (open minus closed is open) and is a neighbourhood of y that misses E, and so shows that y is not in E\'. So U misses E\', and this shows that for every point of X\\E\' we can find a neighbourhood of it that is also inside X\\E\'. So X\\E\' is open and E\' is closed. .

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- Let E be a subset of R n . show that the derived set of E, Eâ€™, is closed Solution Suppose that X is a space such that all singleton subsets of X are closed. (this includes all metric spaces; this condition is known by the name T_1). Then for all subset E of X, E is closed, where A' is the derived set of E. Recall that E' is the set of all x such that every open neighbourhood of x (every open ball, if you are working metrically) intersects E in a point other than x. Proof: Suppose that x is NOT in E'. This means that there is an open neighbourhood U of x, such that U / E is either {x} or empty. Now, every point y of U is also NOT in E': for x this is clear, for all other y we use U{x}, which is open as X has closed singletons (open minus closed is open) and is a neighbourhood of y that misses E, and so shows that y is not in E'. So U misses E', and this shows that for every point of XE' we can find a neighbourhood of it that is also inside XE'. So XE' is open and E' is closed.

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