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I am asked to prove that $\text{aff}(X)$ is closed in the topological sense. I found some posts where people show that it is closed under affine combinations (what it's quite obvious because the definition of affine set).

So i want to know if this prove is equivalent to show that $\text{aff}(X)$ is closed in the topological sense. In other case, I would like to know how to prove it.

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  • $\begingroup$ In general , the claim is false. The vector space of polynomials is an affine subset of the vector space of formal power series, but not closed. $\endgroup$ – Hagen von Eitzen Feb 14 at 7:00
  • $\begingroup$ @HagenvonEitzen In en.wikipedia.org/wiki/Affine_hull it is also a propertie, what surprise me harder $\endgroup$ – Lecter Feb 14 at 7:17
  • $\begingroup$ @Lecter: Wikipedia discusses the case $X \subset \mathbb R^n$. In this case, any affine subspace is closed. In the infinite dimensional case, this is no longer true. It is also worthwhile to rephrase your question as: "Is every affine subspace topologically closed?", since every affine subspace is its own affine hull (obviously). $\endgroup$ – gerw Feb 14 at 7:47

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