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Recall that a face of a convex set $X\subseteq\mathbb{R}^n$ is a convex subset $F$ of $X$ such that every line segment with endpoints in $X$ whose relative interior meets $F$ lies entirely in $F$.

Let $F$ be a face of a (closed) convex set $X\subseteq\mathbb{R}^n$. Let $x,y$ be two points in the relative interior of $F$. Does there exist $\epsilon>0$ such that $(X-x)\cap(\epsilon B)=(X-y)\cap(\epsilon B)$, where $B=\{\,z\in\mathbb{R}^n\mid \|z\|\le1\,\}$ denotes the unit ball and $X-x=\{\,z-x\mid z\in X\,\}$ denotes the set $X$ translated by a vector $x$?

I guess it's true for polyhedra. Is it true for general (closed) convex sets, can you please point me to a proof or give a counter-example?

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  • $\begingroup$ You need to rethink what you mean when you say that "$X$ in the vicinity of $x$ and $y$ looks the same": taking an intersection with a ball $B$ centred on the origin isn't telling you anything about what $X$ looks like near $x$ or near $y$. (My earlier comment about equality stands, but this point is more important.) $\endgroup$ – Rob Arthan Apr 11 at 22:06
  • $\begingroup$ How do you define the face of a general convex set? $\endgroup$ – David M. Apr 11 at 23:36
  • $\begingroup$ I added more details to the question, incl. the definition of a face. $\endgroup$ – Tom Werner Apr 12 at 7:47
  • $\begingroup$ Perhaps I am misunderstanding something- if you consider the closed square. The sides are 1-dimensional faces while the corners are zero dimensional faces. Locally the square cannot be the same at a point in a one dimensional face as it is at a zero dimensional face. Also your definition of "looking the same" doesn't account for rotation, so while I think you would like to say that the circle looks the same at any two points on the boundary, no two points on the boundary will satisfy your equality. $\endgroup$ – Eric Apr 12 at 14:09
  • $\begingroup$ But the points $x,y$ lie in the relative interior of a single face of the square. As for your second question, I removed the (informal) sentence about 'looking the same' to prevent confusion. $\endgroup$ – Tom Werner Apr 12 at 15:46

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