# Show that $F = \{\lambda x|\lambda\in \mathbb{R}_+, x\in K\}$ is closed where $K$ is a compact set.

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?

• What are you calling $C_E \ K$ ? – charmd Nov 7 '18 at 22:14
• Is $R_+$ the positive reals? If so, then it is false. – William Elliot Nov 7 '18 at 22:21
• @CharlesMadeline Complement of the set $K$ in the space $E.$ – model_checker Nov 7 '18 at 22:26
• @WilliamElliot It is the set of all real numbers greater than equal to $0.$ – model_checker Nov 7 '18 at 22:27
• 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 $x \in K$. – Kavi Rama Murthy Nov 7 '18 at 23:35

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 $$y=\lambda x\in F$$.