Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be defined as $f(x) = x$ if $||x||\le1$ and $ \frac{x}{||x||^2}$ if $||x||> 1$. Show that $f$ is continuous.
My attempt: $||x||\le 1$ $\iff x \in B[0,1] = U_1$ and $||x||>1$ $\iff$ $x\in X – B[0,1] = U_2$, it's clear that $f(x) \mid_{x\in U_1}$ is continuous since it is an identity map. Since $||\,.||$ is a continuous function from $X \to\mathbb{R}_{\ge 0}$, $f(x) \mid_{x\in U_2}$ is also continuous. The problem arises when I want to apply the gluing theorem. In the theorem, either all $U_{i}'s$ are open or closed. In this case $U_1$ is closed and $U_2$ is open. So how do I proceed from here?
(Gluing Lemma): Let $X$ and $Y$ be (metric) spaces.
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Let $\{U_i : i \in I\}$ be a family of open sets such that $\cup_{i} U_i = X$. Assume that there exists a continuous function $f_i: U_i \to Y$ for each $i \in I$ with the property that $f_i(x) = f_j(x)$ for all $x \in U_i \cap U_j$ and $i,j \in I$. Then the function $f: X \to Y$ defined by setting $f(x) := f_i(x)$ if $x \in U_i$ is well-defined and continuous on $X$.
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Let $\{A_i : i \in F\}$ be a finite family of closed sets such that $\cup_{i} A_i = X$. Assume that there exists a continuous function $f_i: A_i \to Y$ for each $i \in F$ with the property that $f_i(x) = f_j(x)$ for all $x \in A_i \cap A_j$ and $i,j \in F$. Then the function $f: X \to Y$ defined by setting $f(x) := f_i(x)$ if $x \in A_i$ is well-defined and continuous on $X$ .
Best Answer
Apply 2) to the closed sets $\{x: \|x\| \leq 1\}$ and $\{x: \|x\| \geq 1\}$ and the functions $x$ and $\frac x {\|x\|^{2}}$.