Kdjykuj

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 !

In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.

Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.

The corresponding family of parameters $lambda (t)$ and $k(t)$ are:

$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$

As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.

hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.

If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.

This is wrong. We know $k_{phi(n)}to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.