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Became Hot Network Question
added obvious implicit assumption (false for n<2)
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YCor
YCor
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Let $K$ be a compact subset of $\mathbb R^n$ with $n\ge 2$ (say if you like $n=2$, which is possibly sufficiently representative).

Q: Does there exist a closed simple curve $u:\mathbb S^1\to\mathbb R^n $ such that $K\cup u(\mathbb S^1 )$ is connected?

The set $K$ may have uncountably many connected components, and $u$ has to meet them all. Yet this does not seem a serious obstruction. For instance, the cartesian square of the Cantor set can be connected by some simple self-similar curve (necessarily of infinite length; in fact I think of dimension at least $4/3$), e.g. just connecting suitably the four main square clusters between them by segments, and then iterating.

Let $K$ be a compact subset of $\mathbb R^n$ (say if you like $n=2$, which is possibly sufficiently representative).

Q: Does there exist a closed simple curve $u:\mathbb S^1\to\mathbb R^n $ such that $K\cup u(\mathbb S^1 )$ is connected?

The set $K$ may have uncountably many connected components, and $u$ has to meet them all. Yet this does not seem a serious obstruction. For instance, the cartesian square of the Cantor set can be connected by some simple self-similar curve (necessarily of infinite length; in fact I think of dimension at least $4/3$), e.g. just connecting suitably the four main square clusters between them by segments, and then iterating.

Let $K$ be a compact subset of $\mathbb R^n$ with $n\ge 2$ (say if you like $n=2$, which is possibly sufficiently representative).

Q: Does there exist a closed simple curve $u:\mathbb S^1\to\mathbb R^n $ such that $K\cup u(\mathbb S^1 )$ is connected?

The set $K$ may have uncountably many connected components, and $u$ has to meet them all. Yet this does not seem a serious obstruction. For instance, the cartesian square of the Cantor set can be connected by some simple self-similar curve (necessarily of infinite length; in fact I think of dimension at least $4/3$), e.g. just connecting suitably the four main square clusters between them by segments, and then iterating.

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Pietro Majer
Pietro Majer
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Connecting a compact subset by a simple curve

Let $K$ be a compact subset of $\mathbb R^n$ (say if you like $n=2$, which is possibly sufficiently representative).

Q: Does there exist a closed simple curve $u:\mathbb S^1\to\mathbb R^n $ such that $K\cup u(\mathbb S^1 )$ is connected?

The set $K$ may have uncountably many connected components, and $u$ has to meet them all. Yet this does not seem a serious obstruction. For instance, the cartesian square of the Cantor set can be connected by some simple self-similar curve (necessarily of infinite length; in fact I think of dimension at least $4/3$), e.g. just connecting suitably the four main square clusters between them by segments, and then iterating.