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Let $f:[a,b]\to \mathbb{R}^d$ be a function of class $C^1$. Let us consider the following set: $$ \mathcal{A}:=\bigcup_{x\in [a,b]}\{y\in \mathbb{R}^d, \quad ||y- f(x)||\leq 1/2 \} $$ I think that this set describes a tube of raduis $1/2$ that is centered around the function $f$. Now the goal is to prove that $\mathcal{A}$ is compact set of $\mathbb{R}^d$. To do so, I thought of using the sequential definition: Let $(y_n)_n$ be a sequence that belongs to $\mathcal{A}$. Therefore, there exists $x_n \in [a,b]$ such that $$ y_n \in \{y \in \mathbb{R}^d, \quad ||y-f(x_n)||\leq 1/2\}. $$ Now, from the compactness of $[a,b]$ then, $(x_n)_n$ has a subsequence $(x_{\phi(n)})_n$ that converges towards $x^* \in [a,b]$. This implies that we have $$ y_{\phi(n)} \in \{y \in \mathbb{R}^d, \quad ||y-f(x_{\phi(n)})||\leq 1/2\}. $$ On the other hand, one has $$ ||y_{\phi(n)}-f(x^*)||\leq ||y_{\phi(n)}-f(x_{\phi(n)})|| + ||f(x_{\phi(n)})- f(x^*)||$$ Moreover, for all $\epsilon>0$ then there exists $N\in \mathbb{N}^*$ such that $$ ||f(x_{\phi(n)})- f(x^*)|| \leq \epsilon, $$ for all $n\geq N$. this means that $$ ||y_{\phi(n)}-f(x^*)||\leq 1/2 + \epsilon$$ for all $n\geq N$ i.e. $$ y_{\phi(n)}\in \bigcap_{\epsilon>0} \{y\in \mathbb{R}^d, \quad ||y-f(x^*)||\leq 1/2+\epsilon\} $$ for all $n\geq N$. But I cannot deduce why it necessarly converges in $\mathcal{A}$.

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The Heine-Borel theorem tells us that it is sufficient to check that $\mathcal{A}$ is closed and bounded. Rewrite $\mathcal{A}$ as: $$ \mathcal{A} = \{y \in \mathbb{R}^d: \exists x \in [a, b], \lVert y - f(x) \rVert \le 1/2\} $$

  • Closedness: Let $y$ be an accumulation point of $\mathcal{A}$. Then there exists $(x_n) \subseteq [a, b], (y_n) \subseteq \mathcal{A}$ such that $$ \lVert y_n - f(x_n) \rVert \le \dfrac{1}{2} \ \forall n, \text{ and } \lVert y_n - y \rVert \rightarrow 0 \text{ as } n \rightarrow \infty $$ The compactness of $[a,b]$ implies, there exists a subsequence $(x_{n_k})$ of $(x_n)$ converges to some $x_0 \in [a, b]$. Then, $$ \begin{align*}\lVert y - f(x_0) \rVert &\le \lVert y - y_{n_k} \rVert + \lVert y_{n_k} - f(x_{n_k}) \rVert + \lVert f(x_{n_k}) - f(x_0) \rVert \ \forall k \in \mathbb{N}\\ &\le \lVert y - y_{n_k} \rVert + \lVert f(x_{n_k}) - f(x_0) \rVert + \dfrac{1}{2} \end{align*} $$ which implies $\lVert y - f(x_0) \rVert \le 1/2$ by the continuity of $f$. Thus $y \in \mathcal{A}$, i.e $\mathcal{A}$ is closed.

  • Boundedness: Since $f$ is continuous, $f([a, b])$ is bounded, which means, there exists $M > 0$ such that $$ \lVert f(x) \rVert \le M \ \forall x \in [a, b] $$ For $y \in \mathcal{A}$, there exists $x \in [a, b], \lVert y - f(x) \rVert \le 1/2$. Using the triangle inequality, we have $$ \lVert y \rVert \le \lVert y - f(x) \rVert + \lVert f(x) \rVert \le M + \dfrac{1}{2} $$ Thus, $\mathcal{A}$ is bounded.

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