\(\ZZ\)
\(\ts{1}\)
\(\ts{p \in \ZZ^{\geq 0} \st p\text{ is prime}}\)
\(\ts{ {1\over n} \st n\in \ZZ^{\geq 0}}\)
\(\ts{ {1\over n} \st n\in \ZZ^{\geq 0}} \union \ts{0}\)
Prove that \(\RR^n\) is not homeomorphic to \(\RR\) for any \(n\geq 2\).
Is it true that the closure of a product is the product of the closures?
Solution
1. No operation or property to determine is specified for the first five displayed subsets.
2. For every \(n\ge2\), \(\mathbb R^n\) is not homeomorphic to \(\mathbb R\).
If \(h:\mathbb R^n\to\mathbb R\) were a homeomorphism, then deleting any point \(x\) would give a homeomorphism \(\mathbb R^n\setminus\{x\}\cong\mathbb R\setminus\{h(x)\}\). For \(n\ge2\), \(\mathbb R^n\setminus\{x\}\) is path-connected, while \(\mathbb R\setminus\{h(x)\}\) has two connected components, a contradiction.
3. For arbitrary subsets \(A\subseteq X\) and \(B\subseteq Y\),\[\boxed{\overline{A\times B}^{\,X\times Y}=\overline A^{\,X}\times\overline B^{\,Y}.}\]
If \((x,y)\) lies in the left closure, continuity of the projections gives \(x\in\overline A\) and \(y\in\overline B\). Conversely, if \(x\in\overline A\) and \(y\in\overline B\), every basic neighborhood \(U\times V\) of \((x,y)\) meets \(A\times B\) because \(U\cap A\ne\varnothing\) and \(V\cap B\ne\varnothing\).