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Let ${\mathcal A} \subset {\mathbb R}^{n}$ be a path-connected set, and ${\mathcal B} \subset {\mathbb R}^{n}$ be a linear subspace of ${\mathbb R}^{n}$. Is ${\mathcal A} \bigcap {\mathcal B}$ path-connected?

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  • $\begingroup$ it need not be connected $\endgroup$ – Mirko Aug 13 at 17:34
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    $\begingroup$ No. The intersection of an annulus and a line in R^2 is not connected. $\endgroup$ – Leo Aug 13 at 17:34
  • $\begingroup$ [+1] Thanks a lot! A perfect example! @Leo $\endgroup$ – Ryan Aug 13 at 17:36
  • $\begingroup$ I'm voting to close this question as off-topic because it is solved directly by the comments from Leo. $\endgroup$ – Ryan Aug 13 at 17:38
  • $\begingroup$ @Ryan what did you try? $\endgroup$ – Mirko Aug 13 at 17:47
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There are easy examples (as in the comments) when the intersection need not be connected, so in particular it is not path-connected.

There is also a general method to come up with a path-connected example in a higher dimension, given one that is not path-connected in a lower dimension, although it may be connected. (This perhaps makes it slightly more interesting...)

For example, take a subset $P$ of the plane $\Bbb R^2$ that is connected, but not path connected. Take an extra point $z$ that is in $\Bbb R^3$ but not in the subspace $\Bbb R^2$ that contains $P$. Connect each point $p\in P$ with a straight line segment to $z$, and let $C$ be the union of all such line segments. $C$ is called a cone, and is clearly path-connected, yet its intersection with the subspace $\Bbb R^2$ is $P$, which is not path-connected.

Of course, if ${\mathcal B}={\mathbb R}^{n}$ then ${\mathcal A} \bigcap {\mathcal B} ={\mathcal A}$ which is path-connected :)

But as long as ${\mathcal B}\not={\mathbb R}^{n}$ and $dim{\mathcal B}\ge1$ then we could take any subset $P$ of ${\mathcal B}$ that is not path-connected, and a point $z\in\mathbb R^n\setminus{\mathcal B}$, and the cone $C$ that we obtain by taking all line segments connecting points $p\in P$ with $z$. The cone is always path-connected, and $C\cap{\mathcal B}=P$ which is not path-connected. (As long as $dim{\mathcal B}\ge2$ we may also take $P$ that is connected, but not path-connected.)

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No. The intersection of an annulus and a line in R^2 is not necessarily connected.

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  • $\begingroup$ I agree (with what I believe you mean, such an intersection need not be connected), but just out of nitpicking, sometimes such an intersection may be connected. It is certainly not connected, if the line goes through the origin (as it should, being a linear subspace), and if the annulus is centered at the origin (which is the most natural choice and what one would think of upon reading your answer). But if the line is the $x$-axis, and the annulus is centered at $(0,-2)$ formed of circles with radii $1$ and $3$, then the intersection happens to be connected. $\endgroup$ – Mirko Aug 13 at 18:08
  • $\begingroup$ I added "necessarily". $\endgroup$ – Leo Aug 13 at 18:57

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