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How to show the following:

Let $C$ be a convex subset of $\mathbb R^d$. Then $\operatorname{int} C \neq \emptyset$ if and only if $\operatorname{aff} C = \mathbb R^d$ where $\operatorname{aff} C$ is the smallest affine space containing convex $C$.

Thanks a lot!

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This is plainly wrong as posted here. The filled (d-1)-dimensional unit circle $B_\varepsilon^{d-1}(0)\subset\mathbb{R}^d$ is convex. The smallest affine space (not set, becuase that is not a thing) containing it is $\mathbb{R}^{d-1}$ – CBenni Jan 29 '13 at 8:36
how can be a circle convex??? i guess u r plainly wrong – Salih Ucan Jan 29 '13 at 8:40
I di not now what your definition of convex would be, but mine is And you will not receive any help from me if you start becoming harsh. On top of that, three question marks are not better than one and correct grammar+spelling help you out, even on the internet. – CBenni Jan 29 '13 at 8:43
You're both wrong, methinks. @CBenni said filled circle, also known as a disk. It is convex. On the other hand, I really don't know what a $(d-1)$-dimensional circle is, whether filled or not. But if we assume a $(d-1)$-dimensional ball, it has an empty interior viewed as a subset of $\mathbb{R}^d$, so it is not a counterexample. In fact, the result is true as stated … – Harald Hanche-Olsen Jan 29 '13 at 8:51
up vote 2 down vote accepted

It can be useful to realize that you can shift $C$ around without changing either its convexity, nor the (non-)emptiness of its interior, nor the property of it affinely spanning $\mathbb{R}$. So we might as well assume $0\in C$, in which case the affine span equals the linear span.

If $C$ has a nonempty interior, we can even assume that $0$ is an interior point. in which case the natural basis of $\mathbb{R}^n$, suitably scaled down in size, will lie inside $C$, so $C$ spans all of $\mathbb{R}^n$.

On the other hand, assume $C$ spans $\mathbb{R}^n$. Then $C$ contains a basis of $\mathbb{R}^n$. The interior of the simplex with corners in the origin and those basis vectors is contained in $C$, which therefore has nonempty interior.

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