Bounding diameter of the arc of a closed curve I was reading chapter 4 of Colding and Minicozzi's A Course in Minimal surfaces and I came across a statement in the proof of Lemma 4.14:
Suppose $\Gamma\subset\mathbb{R}^3$ is a simple closed curve of finite arc length. Then for any sufficiently small $\epsilon>0$ there exists $d>0$ such that if $p,q\in\Gamma$ with $0<|p-q|<d$, then $\Gamma\setminus\{p,q\}$ has exactly one component with diameter $<\epsilon$.
I think I know how to prove something like this assuming that $\Gamma$ is $C^2$ (or whatever condition that leads to the existence tubular neighbourhoods), but I'm not sure how to prove this when the curve is simply finite arc-length.
Edit:
Changed ''for any $\epsilon>0$'' to ''for any sufficiently small $\epsilon>0$''.
 A: Proposition. Suppose that $(X,d)$ is a metric space homeomorphic to the circle $S^1$. Then for every $\epsilon>0$ there exists $\delta>0$ such that for all $x,y\in X$ satisfying $d(x,y)<\delta$, the diameter of one of the components of $X-\{x,y\}$ is $<\epsilon$.
Proof. I fix a homeomorphism $f: S^1\to X$. Recall that $f$ is uniformly continuous (in view of compactness of $S^1$). Suppose that the proposition fails. Then there exists $\epsilon>0$ and a pair of sequences $x_n=f(s_n), y_n=f(t_n)\in X$ such that:

*

*$\lim_{n\to\infty}d(x_n,y_n)=0$.


*Both components of $X-\{x_n,y_n\}$ have diameter $\ge \epsilon$.
In view of uniform continuity of $f$, after passing to a subsequence the sequence, $\lim_{n\to\infty} s_n=t= \lim_{n\to\infty} t_n$. Let $a_n$ denote the shorter of the two arcs, components of $S^1 -\{s_n, t_n\}$. Then $diam (a_n)\to 0$  as $n\to \infty$.
By (uniform) continuity of $f$, it follows that $diam (f(a_n))\to 0$ as $n\to \infty$. This is a contradiction.  qed
Now, if $\epsilon$ is too big, then it can happen that for all $x, y\in X$, both components of $X-\{x,y\}$ have diameter $<\epsilon$. However, in this case, $diam(X)<2\epsilon$. Thus, whenever we take $\epsilon\le diam(X)/2$, at most one component of $X-\{x,y\}$ has diameter $<\epsilon$. Combining this observation and the proposition, one obtains the claim that you are after.
