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Let $f:[0,T] \to \mathbb{R}^n$ be a continuous function, I have the intuition that the following set is compact: $$ A:=\bigcup_{t\in [0,T]}\overline{B}(f(t),1) $$ with $\overline{B}(f(t),1)$ is the closed ball of center $f(t)$ and raduis $1$. But, I do not know how to prove it.

My idea: is that since we are in finite dimension, then, it sufficies to prove that is closed and bounded. It is obvious that it is closed (as the union of closed sets). But I am struggling with the compactness. I think of the following argument $$ A\subset \overline{B}(0, 2M) $$ where $M:=\sup_{t\in[0,T]}\|f(t)\|$.

hanava331
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    Boundedness is actually easy, as $f$ has bounded range, so extending by radius 1 still results in a bounded region. However, your statement that $A$ is closed because it is a union of closed sets is wrong. Singletons are closed, so literally every set is a union of closed sets. Obviously not all sets are closed. – David Gao Feb 13 '24 at 08:12
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    Regardless, $A$ is compact, because it is exactly the set of all points with distance less than or equal to 1 from the compact set $f([0, T])$. – David Gao Feb 13 '24 at 08:15
  • @DavidGao Thank you for your reply. And how to prove the closedness? – hanava331 Feb 13 '24 at 08:22
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    @hanava331 the function $x\mapsto d(x,f([0,T]))$ is continuous and, as David pointed out in a previous comment, the set you're interested in is the preimage of the closed set $[0,1]$ through this function – Alessandro Codenotti Feb 13 '24 at 08:42
  • Another way to see it is that the multivalued function $F(t)=\overline{B}(f(t),1)$ is upper semi continuous. And thus image of compact is compact, see this: https://math.stackexchange.com/questions/1895973/proving-image-of-compact-set-via-upper-semicontinuous-set-valued-map-is-compact – freakish Feb 13 '24 at 09:17

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Your question can be generalized as follows:

Let $C \subset \mathbb R^n$ be compact. Show that $$C' = \bigcup_{x \in C} \overline B(x,1)$$ is compact.

In fact $C'$ is bounded. An upper bound is $M + 1$ with $M = \sup_{x \in C} \lVert x \rVert$. Taking $2M$ is not a good idea because $M < 1$ is possible.

It remains to prove that $C'$ is closed. Note that unions of closed sets are in general not closed, thus a better argument is needed.

Let $(x'_n)$ be a sequence in $C'$ converging to some $\xi' \in \mathbb R^n$. We have to show that $\xi' \in C'$. Choose $x_n \in C$ such that $x'_n \in \overline B(x_n,1)$. Since $C$ is compact, $(x_n)$ has a subsequence $(x_{n_k})$ converging to some $\xi \in C$. Clearly $x'_{n_k} \to \xi'$. For each $m \in \mathbb N$ there exists $k_0(m)$ such that $\lVert \xi -x_{n_k} \rVert < 1/m$ and $\lVert \xi' -x'_{n_k} \rVert < 1/m$ for $k \ge k_0(m)$. For each $m$ we get with $k \ge k_0(m)$ $$\lVert \xi' - \xi \rVert \le \lVert \xi' -x'_{n_k} \rVert + \lVert x'_{n_k} -x_{n_k} \rVert + \lVert x_{n_k} - \xi \rVert < 1/m + 1 + 1/m = 1 + 2/m.$$ This is possible only when $\lVert \xi' - \xi \rVert \le 1$, thus $\xi' \in \overline B(\xi,1) \subset C'$.

Paul Frost
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