I have a proof of the following **false** fact : 

> 
> Let $E$ be normed vector space. Let $K \subset E$ be a compact set. Then the set $B = \{\lambda x \mid \lambda \in \mathbb{R}^+, x \in K \}$ is closed (where $\mathbb{R}^+$ are the positive real numbers including $0$).
> 
> 
> 

This fact is true when $0 \not \in K$ yet it can be false when $0 \in K$. For example by taking the semi-circle in the plane centered at $(1,0)$ of radius $1$.

So I made a proof a of this fact. My proof is thus obviously **false** yet I don't see where the mistake is : 

Let $(\lambda\_n k\_n)$ be a sequence in $B$ which converges to a vector $x \in E$. We want to prove that $x \in B$. 

Since $K$ is compact there is $\phi : \mathbb{N} \to \mathbb{N}$ strictly increasing such that $(k\_{\phi(n)})$ converges to a vector $k \in K$. If $ k = 0$ then the sequence $(\lambda\_n k\_n)$ converges to $0 \in B$ and we are done. So we can suppose $k \ne 0$.

Since the sequence $(\lambda\_n k\_n)$ converges to $x$ we must have $k \in span \{ x \}$. So there is $\mu \in \mathbb{R}^\*$ such that $k = \mu x$. From here we can deduce that the sequence $(\lambda\_n)$ necessarily converges to $\frac{1}{\mu}$. Yet since $\mathbb{R}^+$ is closed the sequence $(\lambda\_n)$ converges to a positive real number, so $\frac{1}{\mu} \geq 0$ so $\mu \geq 0$. 
So the sequence $(\lambda\_n k\_n)$ converges to the vector $\frac{1}{\mu} k \in B$ since $k \in K$ and $\frac{1}{\mu} \geq 0$. Hence $B$ is closed.

So where am I going wrong here ?

Thank you !