aff S = aff cl S in $R^n$

Question

I need to prove that the set of aff S = aff cl S in $R^n$ under the standard metric.

Definitions :

1. cl S is the closure of S .
2. Affinity hull of S is the smallest affine set that contains S. Notation: aff S is the affine hull of S.

My tentative approach:

-> aff S $\subset$ aff cl S.

We know that S is a subset of the closure of S, therefore, by the definition of affine hull, aff S is a subset of aff cl S.

<- aff cl S $\subset$ aff S.

I am not sure how to do the converse, any help would be appreciated.

Answer 1

This is not, in general, true. For consider the set $X = C^0([0, 1])$ under the $\sup$ norm, and consider $S \subseteq X$, the set of polynomial functions. We see that $S$ is affine, and hence its own affine hull. But the closure of $S$ includes functions which are not in $S$ (such as $f(x) = e^x$), and hence the affine hull of the closure of $S$ also includes these functions. $\DeclareMathOperator{aff}{aff} \DeclareMathOperator{cl}{cl}$

Note: the question has been edited to only concern $S \subseteq \mathbb{R}^n$. In this case, the proof is quite simple. All affine subspaces of $\mathbb{R}^n$ are closed. Therefore, $S \subseteq \cl(S) \subseteq \aff(S)$. Then $\aff(S) \subseteq \aff(\cl(S)) \subseteq \aff(\aff(S)) = \aff(S)$.