In Euclidean space, is the intersection of a path-connected set and a linear subspace still path-connected?

Question

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?

Answer 1

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.)

Answer 2

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