Exercise · ZW-HAUSDORFF-SPACE-006 · ★★☆ Intermediate
In a Hausdorff space \(X\), what can be said about each singleton \(\{x\}\)?
Choose one option.
Show answer
- D. It is closed
Explanation
For each \(y\neq x\) there is an open \(V\ni y\) not containing \(x\); so \(X\setminus\{x\}\) is a union of open sets and \(\{x\}\) is closed. In the usual \(\mathbb{R}\), \(\{x\}\) is not open.
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
- ZW-HAUSDORFF-SPACE-001 Which of these topologies on X=\1,2,3\ is Hausdorff?
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- ZW-HAUSDORFF-SPACE-005 If X is Hausdorff and A⊆ X, the subspace A is:
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- ZW-HAUSDORFF-SPACE-008 Let f:X→ Y be continuous and injective, with Y Hausdorff. What follows about X?
- ZW-HAUSDORFF-SPACE-009 Let X=\a,b\ have the trivial topology and Y any Hausdorff space with two or more points. Is Y× X wi…