Let X=\1,2,3\ with open sets ∅,\1\,\2,3\,X, and Y=\a,b,c\ with open sets ∅,\a\,\a,b\,Y. Both have 4 open sets. Are they homeomorphic?

No: X has two disjoint proper open sets and Y does not

Yes: both have 3 points and 4 open sets

Yes: the bijection 1mapsto a,\ 2mapsto b,\ 3mapsto c is enough

No: they do not have the same number of points

Exercise · ZW-TOPOLOGICAL-HOMEOMORPHISM-009 · ★★★ Advanced

Let \(X=\{1,2,3\}\) with open sets \(\varnothing,\{1\},\{2,3\},X\), and \(Y=\{a,b,c\}\) with open sets \(\varnothing,\{a\},\{a,b\},Y\). Both have 4 open sets. Are they homeomorphic?

Choose one option.

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B. No: \(X\) has two disjoint proper open sets and \(Y\) does not

Explanation

A homeomorphism must send disjoint open sets to disjoint open sets. In \(X\), \(\{1\}\) and \(\{2,3\}\) are disjoint proper open sets; in \(Y\) the proper open sets \(\{a\}\subset\{a,b\}\) are nested. Counting points and open sets is not enough; checking all 6 bijections shows none works.

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Review the concept: Homeomorphism

A bijection \(f\) between topological spaces \(X\) and \(Y\) that is continuous and whose inverse \(f^{-1}\) is also continuous.

\[f:X\to Y\ \text{bijective};\quad f,\ f^{-1}\ \text{continuous}\]

Let \(X=\{1,2,3\}\) with open sets \(\varnothing,\{1\},\{2,3\},X\), and \(Y=\{a,b,c\}\) with open sets \(\varnothing,\{a\},\{a,b\},Y\). Both have 4 open sets. Are they homeomorphic?

Explanation

Review the concept: Homeomorphism

More practice problems: homeomorphism