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Answer

Can Nice Spaces be Broken into Two Equal Parts? Attilio Le Donne Asked in Can Nice Spaces be Broken into Two Equal Parts? by Stephen Watson

Clearly the two questions can be answered if we can find spaces X, with these properties, so that there is a point p in X so that each closed set not containing p has cardinality different from that of X.

For the first question take the metric hedgehog with for each n, \alpha_n spines of length 1/n, where \alpha_n is an increasing sequence of cardinals bigger than 2^\omega.

For the second question, under the assumption \omega_\omega <= 2^\omega, a dense in itself subset of [0,1] so that the cardinality of X \cap [1/(n+1), 1/n] is \omega_n and 0 in X gives a consistent answer.