# affine subset closedness

How to show an affine subset $C$ of $R^d$ is closed (yes convergence in norm sense) I thought the following: Let $x_0$ be an accumulation point $C$ define $C - C + x_0$ then $C-C$ being closed $C - C + x_0$ is also closed. Let $y_n \rightarrow y$ and $x_n \rightarrow x_0$ then $y_n - x_n + x_0$ converges to $y$ which belongs to $C$ so we are done.. But i didnt like my proof..?

Thank you...

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Do you know the fact that all (linear) subspaces of a finite dimensional (real or complex) vector spaces are automatically closed subsets? It follows from the theorem that asserts that all norms on a finite dimensional vector space are equivalent. This would easily solve your question. – Olivier Bégassat Jan 28 '13 at 23:03
Following @OlivierBégassat's comment, choose the Euclidean norm and show that any point $x \in C^C$ has a'perpendicular' distance $\delta>0$ from $C$, hence $B(x,\frac{1}{2}\delta) \cap C = \emptyset$. Hence $C^C$ is open. – copper.hat Jan 28 '13 at 23:07
How do you use the closedness of finite vector space??? – Salih Ucan Jan 28 '13 at 23:20
This is what you are trying to show? – copper.hat Jan 29 '13 at 1:59