showing $\{\alpha u : \alpha\in \mathbb{R}\}$ and $\{w+\beta v:w\in Y, \beta\in\mathbb{R}\}$ are closed sets

Question

Let $X$ be a normed vector space.

How can I show the following?

(1) Let $0\neq u\in X$. Show that $\{\alpha u : \alpha\in \mathbb{R}\}$ is closed.

(2) If $Y$ is a closed vector subspace of $X$ and $v\in X-Y$, then $\{w+\beta v:w\in Y, \beta\in\mathbb{R}\}$ is closed.

Hope you can help! Thank you.

(1) Take a convergent sequence in the set and verify that the limit must lie in a compact subset. — Michael Greinecker, Jan 16 '14 at 23:57
This might help you: Any finite-dimensional Hausdorff vector space over $\Bbb R$ is isomorphic to $\Bbb R^n$ (topologically and algebraically), thus complete. Or show directly that ${αu:α\in\Bbb R}$ is complete. And a complete subset of a Hausdorff space is closed. — Stefan Hamcke, Jan 16 '14 at 23:59
This just came up recently. http://math.stackexchange.com/questions/639570/sum-of-closed-subspaces-of-normed-linear-space/639583#639583 — Disintegrating By Parts, Jan 17 '14 at 04:55

score 2 · Answer 1 · answered Jan 17 '14 at 00:48

(1) Suppose $\|u\|=1$ and $(\alpha_nu)_n$ converges to some $x\in X$. Then $(\alpha_n)$ is bounded because $$|\alpha_n|=\|\alpha_nu\|=\|(\alpha_nu-x)+x\|\le \|\alpha_nu-x\|+\|x\|\,.$$ So, a convergent subsequence $\alpha_{n_k}$ can be chosen, say with limit $\alpha$, but then $\alpha_{n_k}u\overset{k\to\infty}\longrightarrow\alpha u$, so $x$ must be $\alpha u$.

(2) Consider the quotient space $X/Y$ with canonical projection $\phi:X\to X/Y$ sending $x\mapsto [x]$. Since $Y$ is closed, the inherited norm $X/Y\ $ ($\|[x]\|_{X/Y}:=d(x,Y)$) is indeed a norm, and we have $$\phi^{-1}(\{\alpha[v]\mid\alpha\in\Bbb R\})=\{y+\beta v \mid y\in Y,\beta\in\Bbb R\}$$ so this set is the preimage of a closed subspace, hence is closed.

showing $\{\alpha u : \alpha\in \mathbb{R}\}$ and $\{w+\beta v:w\in Y, \beta\in\mathbb{R}\}$ are closed sets

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