Let \(u_1,\ldots,u_k\) be linearly independent vectors in \(\mathbb R^n\), and let \(A\subseteq\mathbb R^k\) be closed. Define \[ S=\left\{\alpha_1u_1+\cdots+\alpha_ku_k:(\alpha_1,\ldots,\alpha_k)\in A\right\}. \] Show that \(S\) is closed in \(\mathbb R^n\).
Solution
Extend \(u_1,\ldots,u_k\) to a basis \(u_1,\ldots,u_n\) of \(\RR^n\), and let \[ U:\RR^n\longrightarrow\RR^n \] be the linear map whose matrix has columns \(u_1,\ldots,u_n\) in the standard basis.
1. The map \(U\) is a homeomorphism of \(\RR^n\).
The vectors \(u_1,\ldots,u_n\) form a basis, so the matrix of \(U\) is invertible. Hence \(U\) and \(U^{-1}\) are linear maps between finite-dimensional normed spaces and are continuous. Thus \(U\) is a homeomorphism.
2. The subset\[ A\times\{0\}\subseteq\RR^k\times\RR^{n-k}=\RR^n \]is closed.
The set \(A\) is closed in \(\RR^k\) by hypothesis, and \(\{0\}\) is closed in \(\RR^{n-k}\). Their product is therefore closed in \(\RR^k\times\RR^{n-k}\).
3. The image of \(A\times\{0\}\) under \(U\) is exactly \(S\).
For \((\alpha_1,\ldots,\alpha_k,0,\ldots,0)\in A\times\{0\}\), the definition of \(U\) gives \[ U(\alpha_1,\ldots,\alpha_k,0,\ldots,0) =\alpha_1u_1+\cdots+\alpha_ku_k. \] As \((\alpha_1,\ldots,\alpha_k)\) ranges over \(A\), these are exactly the elements of \(S\).
4. The set \(S\) is closed in \(\RR^n\).
By step 2, \(A\times\{0\}\) is closed. By step 1, the homeomorphism \(U\) maps closed sets to closed sets. Step 3 identifies its image with \(S\), so \(S\) is closed.
5. Q.E.D.
Step 4 is the required conclusion.