Show that $F = \{\lambda x|\lambda\in \mathbb{R}_+, x\in K\}$ is closed where $K$ is a compact set of a normed vector space $E$ and $0\not \in K.$

I started with a convergent sequence $y_n\in F$ such that $y_n = \lambda_nx_n$ for $\lambda_n\geq 0$ and $x_n\in K.$ Let $||y_n||\to ||y||$ and since $0\not \in K$ there exists $\alpha>0$ such that $B(0,\alpha)\subset C_E K.$ Therefore $\alpha \leq ||x||\leq M$ for some $M>0.$ Thus $\{\lambda_n\}$ is a bounded sequence. Since $K$ is compact we know that there exists a subsequence $\{x_{\phi(n)}\}$ which is convergent. Furthermore the sequence $\{\lambda_n\}$ lives in a compact set in $\mathbb{R}$ and therefore has a convergent subsequence $\{\lambda_{\psi(n)}\}.$ Then $\{y_{\psi(\phi(n))}\}$ is a convergent subsequence in $K.$

Is this proof correct?

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    $\begingroup$ What are you calling $C_E \ K$ ? $\endgroup$ – charmd Nov 7 '18 at 22:14
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    $\begingroup$ Is $R_+$ the positive reals? If so, then it is false. $\endgroup$ – William Elliot Nov 7 '18 at 22:21
  • $\begingroup$ @CharlesMadeline Complement of the set $K$ in the space $E.$ $\endgroup$ – model_checker Nov 7 '18 at 22:26
  • $\begingroup$ @WilliamElliot It is the set of all real numbers greater than equal to $0.$ $\endgroup$ – model_checker Nov 7 '18 at 22:27
  • $\begingroup$ The last sentence says nothi ng. It is something you already knew in the beginning of the proof. You have to show that $\lim y_n =\lambda x$ for some $\lambda \geq 0$ and $x \in K$. $\endgroup$ – Kavi Rama Murthy Nov 7 '18 at 23:35

We can correct your proof as follows.

Let $y\in\overline{F}$ be any point. There exists a sequence $\{y_n\}$ of elements of $F$ convergent to $y$. For each $n$ pick $\lambda_n\geq 0$ and $x_n\in K$ such that $y_n = \lambda_nx_n$. Since $K$ is compact, a sequence $\{x_n\}$ has a convergent subsequence. Without loss of generality we can suppose that the sequence $\{x_n\}$ converges to a point $x\in K$. Since $0\not \in K$, there exists $\alpha>0$ such that $\|z\|\ge\alpha$ for each $z\in K$. Therefore $\|y_n\|=\lambda_n\|x_n\|\ge \lambda_n\alpha$ for each $n$. Since the sequence $\{y_n\}$ converges, a sequence $\{\|y_n\|\}$ is bounded, so a sequence $\{\lambda_n\}$ is bounded too. So it contains a subsequence $\{\lambda_n:n\in I\}$ convergent to $\lambda\in\Bbb R$. Then a sequence $\{\lambda_n x_n:n\in I\}$ converges to a point $y=\lambda x\in F$.


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