T1 not implies Hausdorff

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This article gives the statement and possibly, proof, of a non-implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., T1 space (?)) need not satisfy the second topological space property (i.e., Hausdorff space (?))
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Statement

A T1 space need not be a Hausdorff space.

Related facts

Proof

Example of cofinite topology

Consider a countable set, say the set of natural numbers, equipped with the cofinite topology. In the cofinite topology, the nonempty open subsets are precisely the cofinite subsets -- the subsets whose complement is finite. With this topology:

  • The space is : Every point is closed, because its complement is an open subset. Equivalently, if are two natural numbers, the complement of is an open subset containing but not .
  • The space is not Hausdorff: it is not possible to separate two points with disjoint open subsets, because any two nonempty open subsets have an infinite intersection. This is because the union of their complements is a union of two finite subsets, and hence is not the whole space. (Another way of phrasing this is that the space is an irreducible space and hence, since it has more than one point, it is not a Hausdorff space).

Example of line with two origins

Further information: line with two origins

Consider the line with two origins. This is the real line with two copies of the point . Alternatively, it is the quotient of two copies of the real line where we identify all the nonzero points on the first real line with the corresponding nonzero point on the second real line.

This is a space, but is not Hausdorff, because the two origins cannot be separated by disjoint open subsets.