# compactness of a set where am I going wrong

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 !

• I guess you mean $B$ to be the union of the $\lambda K$ rather than the set of them? – Eric Wofsey Mar 15 at 22:41
• @EricWofsey By this notation I mean the set of $\lambda x$ for all $x \in K$ and $\lambda \geq 0$, so my notation is wrong ? – mouargmouarg Mar 15 at 22:43
• Yes, your notation meant that $B$ is the set whose elements are the sets $\lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$). – Eric Wofsey Mar 15 at 22:45
• @EricWofsey You are right thank you. – mouargmouarg Mar 15 at 22:49
• @SantanaAfton Euh, he/she is right $\mathbb{R}^+$ is indeed closed – Thinking Mar 16 at 7:46

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.

• You are right, it's more clear to me what's happening now. Thank you. – mouargmouarg Mar 15 at 22:56

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.

• Oh, thank you. I should have spotted that... For example by taking : $k_n = \frac{1}{n}$ and $\lambda_n = n$ – mouargmouarg Mar 15 at 22:49