Exercise · ZW-HAUSDORFF-SPACE-008 · ★★☆ Intermediate
Let \(f:X\to Y\) be continuous and injective, with \(Y\) Hausdorff. What follows about \(X\)?
Choose one option.
Show answer
- C. It is Hausdorff
Explanation
If \(x\neq y\) then \(f(x)\neq f(y)\) and there are disjoint open sets \(U\ni f(x)\), \(V\ni f(y)\). Their preimages \(f^{-1}(U)\) and \(f^{-1}(V)\) are disjoint open sets separating \(x\) and \(y\).
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 \(V\ni y\).
\[x\neq y\ \Rightarrow\ \exists\,U,V\ \text{open}:\ x\in U,\ y\in V,\ U\cap V=\varnothing\]
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