Let \(A\) be a closed subset of the normal space \(X\text{.}\)
Any continuous function \(f\colon A\rightarrow [a,b]\) can be extended to a continuous function \(\widetilde{f}\colon X\rightarrow [a,b]\text{.}\)
Any continuous function \(f\colon A\rightarrow \R\) can be extended to a continuous function \(\widetilde{f}\colon X\rightarrow \R\text{.}\)
Proof of (1).
First we show that for any continuous function \(f\colon A\rightarrow [-r,r]\text{,}\) we can find a continuous function \(g\colon X\rightarrow [-r/3, r/3]\) such that (i) \(\abs{g(x)}\leq \frac{1}{3}r\) and (ii) \(\abs{g(a)-f(a)}\leq \frac{2}{3}r\) for all \(a\in A\text{.}\) To do so, subdivide \([-r,r]\) as
define \(B=f^{-1}([-r,-r/3])\) and \(C=f^{-1}([r/3,r])\text{,}\) and, using Urysohn’s lemma, choose a continuous function \(g\colon X\rightarrow [-r/3,r/3]\) such that \(g(B)=\{-r/3\}\) and \(g(C)=\{r/3\}\text{.}\) It is easily verified that this \(g\) does the trick.
Now assume we have a continuous function \(f\colon A\rightarrow [a,b]\text{.}\) Letting \(\phi\colon [a,b]\rightarrow [-1,1]\text{,}\) we see that \(f\) extends continuously to \(X\) if and only if \(\phi\circ f\) extends continuously to \(X\text{.}\) Thus, we may assume \(f\colon X\rightarrow [-1,1]\text{.}\) We can build a sequence of continuous functions \((g_n)_{n\in \Z_+}\) satisfying the following properties:
\(\abs{g_n(x)}\leq \frac{1}{3}\left(\frac{2}{3}\right)^{n-1}\) for all \(x\in X\text{.}\)
\(\abs{f(a)-\sum_{k=1}^ng_k(a)}\leq \left(\frac{2}{3}\right)^n\) for all \(a\in A\text{.}\)
It follows from some convergence of infinite sums arguments that \(\widetilde{f}(x)=\sum_{k=1}^\infty g_k(x)\) is a well-defined continuous function from \(X\) to \([-1,1]\) satisfying \(\widetilde{f}(a)=f(a)\) for all \(a\in A\text{.}\) (See Munkres for these details. Basically just the Weierstrass M-test, if that is familiar to you.)
How dow we build this sequence? From our first paragraph, we can pick \(g_1\) to be any function satisfying \(\abs{g(x)}\leq \frac{1}{3}\) and \(\abs{f(a)-g_1(a)}\leq \frac{2}{3}\text{.}\) Next assume we have constructed \(g_k\) as specified for all \(1\leq k\leq n\text{.}\) Apply the same reasoning to the function \(h=f-g_1-g_2-\cdots -g_n\text{,}\) which by assumption maps to \([-(2/3)^n, (2/3)^n]\) to find \(g_{n+1}\text{.}\) By the recursion principle we get a sequence \((g_n)\) satisfying (i) and (ii) for all \(n\text{.}\)
Proof of (2).
Let \(\phi\colon \R\rightarrow (-1,1)\) be a homeomorphism. A continuous function \(f\colon A\rightarrow \R\) extends continuously to a function \(\widetilde{f}\colon X\rightarrow \R\) if and only if \(h=\phi\circ f\colon A\rightarrow (-1,1)\) extends continuously to a function \(\widetilde{h}\colon X\rightarrow (-1,1)\text{.}\) Thus we may assume \(f\colon A\rightarrow (-1,1)\text{.}\) Since \((-1,1)\subseteq [-1,1]\text{,}\) by (1) there is a function \(\widetilde{f}\colon X\rightarrow [-1,1]\text{.}\) Let \(B=\widetilde{f}^{-1}(\{-1,1\})\text{.}\) If \(B\) is nonempty, then \(\widetilde{f}\colon X\rightarrow (-1,1)\text{,}\) as desired. Otherwise, since \(B\) is closed and \(A\cap B=\emptyset\text{,}\) there is a continuous function \(g\colon X\rightarrow [0,1]\) such that \(g(A)=\{1\}\) and \(g(B)=\{0\}\text{.}\) The function \(h=g\widetilde{f}\) is an extension of \(f\) on \(A\) that maps into \((-1,1)\text{,}\) as desired.
Topological specimen19.Space-filling curve.
We will show that there is a continuous surjective function \(\phi\colon [0,1]\rightarrow [0,1]\times[0,1]\text{.}\) This \(\phi\) is thus a path or curve, whose image is all of \([0,1]\times [0,1]\text{.}\) This called a space-filling curve, or Peano curve.
Proof.
Definition1.21.2.Partition of unity.
Let \(Y\subseteq \R\text{.}\) The support \(\operatorname{supp} f\) of a continuous function \(f\colon X\rightarrow Y\) is defined as \(\operatorname{supp} f=\overline{f^{-1}(\{0\})}\text{.}\)
A partition of unity dominated by the open cover \(X=\bigcup_{k=1}^nU_k\) is a collection of continuous functions \(\{\phi_k\colon X\rightarrow [0,1]\}_{k=1}^n\) satisfying (i) \(\operatorname{supp} \phi_k\subseteq U_k\) for all \(1\leq k\leq n\text{,}\) and (ii) \(\sum_{k=1}^n\phi_k(x)=1\) for all \(x\in X\text{.}\)
Theorem1.21.3.Partition of unity (finite).
Let \(X\) be a normal space. Given any finite open covering \(\mathcal{U}=\{U_k\}_{k=1}^n\) of \(X\text{,}\) there is a partition of unity dominated by \(\mathcal{U}\text{.}\)
Proof.
Not covered in WQ23 course. See Munkres if interested.
Corollary1.21.4.Compact manifold embedding.
Let \(M\) be a compact \(m\)-manifold. There is an embedding \(f\colon M\rightarrow \R^N\) for some \(N\geq 1\text{.}\)
Proof.
Not covered in WQ23 course. See Munkres if interested.
