why is an annulus close to it's boundary when it's boundary curves are close?

Question

This is the motivating question for the rather vague question here: when is the region bounded by a Jordan curve "skinny"?

Suppose we are given two Jordan curves in the plane, one inside the other and each contained in an epsilon neighborhood of the other. How can we conclude that the annular region between the two curves is itself contained in the two epsilon neighborhoods of the original curves?

Note: by epsilon neighborhood of a curve I mean the set of all points which lie within a distance epsilon from some point in the image of the curve.

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Follow up: Since the statement above is false (see Hagen's answer) we might ask whether it is true if we assume point-wise closeness of chosen parametrizations of the curves. This is answered in the affirmative by user7...

Answer 1

We can't. Consider an arbitrarily large closed disk. At two points of its circumference let two thin strips emerge that first wind araound back and forth very close to the disk boundary and then meet. The contour arount this figure consists of two Jordan curves making a counterexample.

Answer 2

I think this may be very similar to the counterexample that Hagen von Eitzen described in words, but I'm not entirely sure. So here's a fooplot.

If you assume that $|f(\theta)-g(\theta)|< 2\epsilon$ for all $\theta$ (where $f$ and $g$ are the two curves), then the conclusion is true. Indeed, suppose that the disk $|z-a|< \epsilon$ is contained in the annulus. We may and do assume $a=0$. Then the winding numbers of $f$ and $g$ around $0$ are different: one is $0$, the other is $1$. On the other hand, I claim that $$f_t(\theta)=(1-t)f(\theta)+tg(\theta)$$ is a homotopy between $f$ and $g$ in $\mathbb R^2\setminus \{0\}$.

Indeed, if $f_t(\theta)=0$ for some $\theta$, then the points $f(\theta),0, g(\theta)$ lie on a line, in the order listed. Hence $|f(\theta)-g(\theta)|=|f(\theta)|+|g(\theta)|\ge 2\epsilon$, which is a contradiction.