Is $U=$ {$ax, a \in R$} closed subset of normed infinite dimensional vector space?

Asked Oct 05 '23 at 14:44

Active Oct 05 '23 at 14:44

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Let X be infinite dimensional normed vector space over $R$. Let $U=$ {$ax, a \in R$} $V=$ {$bx, b \in R$}. Is $U$ closed set? Is $U+V$ closed set?

Sequence from $U$ is $a_nx$ and if it is convergent then it intuitively has to be convergent to $ax$ where $a$ is a limit of $a_n$ so the limit is still in $U$ however I can't prove formally that $a_n$ converges to $y \in X$ implies $y=kx$ for some $k \in R$.

Is $U+V$ closed? $U+V=$ {$ax+by : a,b \in R$}. (Also very important for me question)

asked Oct 05 '23 at 14:44

romperextremeabuser

6

Does this answer your question? Finite-dimensional subspace normed vector space is closed or this? https://math.stackexchange.com/questions/1764798 – Anne Bauval Oct 05 '23 at 14:47
No my question is about Infinite dimensional space – romperextremeabuser Oct 05 '23 at 15:01
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So are these posts. Your question is about subspaces of dimension 0,1, or 2 in that infinite dimensional space. This is a particular case of these posts. – Anne Bauval Oct 05 '23 at 15:02

Is $U=$ {$ax, a \in R$} closed subset of normed infinite dimensional vector space?

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