Is there an easy proof that every closed curve is contained in a ball?

Question

I actually wanted to generalize this to higher dimensions (and prove something akin to the statement "every $\bf{n}$ dimensional closed manifold is contained in an $\bf{n}$ dimensional ball), but I'm very new at this yet so I don't really have a clue. It seems obvious enough (I mean, take any closed planar curve and there exists a circle in which it is contained... the same goes for closed surfaces, I can always find a sphere big enough), but I see no clear path on proving it.

Answer 1

I assume closed means compact without boundary and your manifold is inside euclidean space. Since compact sets of a metric space are bounded, you have your ball for free.

Answer 2

How is your definition of closed curve? If it just a continuous image of $S^1$ then it is compact, so bounded in $\mathbb{R}^n$

Answer 3

Maybe too late, but this is a quite simple explanation:

Consider a closed curve as a function: $f:[0,1]\to\mathbb{R}^n$.

Next, this function must be continuous on the entire domain $[0,1]$ (since it's a curve), and the $f(0) = f(1)$ equality must hold (since it's closed, having no endpoints).

Continuousness in $\mathbb{R}^n$ means that for each $\varepsilon>0$ exists such $\delta>0$ that if $|x_2-x_1|<\varepsilon$, then $D(f(x_2),f(x_1)) < \delta$ in some metric (distance function $D$) defined in $\mathbb{R}^n$.

Of course, this must hold for $\varepsilon=1$, so we get that for any $x_1, x_2 \in [0,1]$ must be $D(f(x_2),f(x_1)) < D_{max}$, where $D_{max}$ is $\delta$ from above, defined for $\varepsilon=1$ (since $|x_2-x_1|<1$ holds for any $x_1, x_2 \in [0,1]$, except 0 and 1 themselves, but we know that $f(0)=f(1)$, so $D(f(1), f(0))=0$).

So, recalling definition of a ball $$B_d(c)=\{x \in \mathbb{R}^n | D(c,x)<d\}$$ (a ball with center $c$ and radius $d$), we can fit our curve in a ball (for example, taking $c=f(0)$ and $d=D_{max}$)