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.


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.

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    $\begingroup$ I think this answer could be improved: Is it by definition? By a theorem? Etc. $\endgroup$ – Daniel R. Collins Sep 11 '18 at 5:38

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

  • $\begingroup$ Hello to the anonymous downvoter who failed to tell me what his problem with my answer is. That way (s)he missed the oppurtunity to tell me how to improve my answers. $\endgroup$ – Thomas Sep 16 '18 at 20:26

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 $d=D_{max}$)


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