Exercise · ZW-HAUSDORFF-SPACE-003 · ★☆☆ Basic

Which of these topologies on \(\mathbb{R}\) is not Hausdorff?

Choose one option.

Show answer
  • A. The cofinite one (open sets: \(\varnothing\) and complements of finite sets)

Explanation

In the cofinite topology any two nonempty open sets meet, since their complements are finite and \(\mathbb{R}\) is infinite. The other three do separate points: with intervals, singletons or \([x,m)\) and \([m,y)\).

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Review the concept: Hausdorff space

A topological space \(X\) is Hausdorff if for every pair of distinct points \(x\) and \(y\) there exist disjoint open sets \(U\ni x\) and \(V\ni y\).

\[x\neq y\ \Rightarrow\ \exists\,U,V\ \text{open}:\ x\in U,\ y\in V,\ U\cap V=\varnothing\]

More practice problems: hausdorff space

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