# JCT - Boundaries of the components of the complement of a Jordan curve II

Scott E. Brodie, MD, PhD

Icahn School of Medicine at Mount Sinai

New York

Recall: The boundary of each component of the complement is a subset of the Jordan curve.

In fact, the boundary of any component of the complement is the entire Jordan curve.

Suppose otherwise. Consider $x,$ $y$ in distinct components of a Jordan curve, respectively.

Recall: If a set does not separate $x,\;y,$ then no subset of it separates $x,\;y.$

Select a component of the complement of the Jordan curve which contains $x\;$ or $y.\;$

Since the boundary of a component is closed, if it omits a point of the Jordan curve, the inverse image of the boundary must therefore omit the corresponding point in the circle which is the pre-image of the Jordan curve.

But since the pre-image of the boundary is also closed, it must omit an open interval surrounding the omitted point.

But by the continuity of the

*inverse*of the mapping which specifies the Jordan curve, the image of this open interval is an omitted*open arc*in the Jordan curve, and its complement, which contains the boundary, must be a (closed) Jordan arc.So ...

Since "boundaries separate," the boundary of the selected component separates $x,$ $y.$

But no Jordan arc separates the plane, so no subset of a Jordan arc, such as the boundary, can separate $x,$ $y.$

This contradiction implies that the boundary must be the entire Jordan curve.

Since the component was arbitrary, the (entire) Jordan curve must be the boundary of each component.

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