7
$\begingroup$

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 !

$\endgroup$
  • $\begingroup$ I guess you mean $B$ to be the union of the $\lambda K$ rather than the set of them? $\endgroup$ – Eric Wofsey Mar 15 at 22:41
  • $\begingroup$ @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 ? $\endgroup$ – mouargmouarg Mar 15 at 22:43
  • $\begingroup$ 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 $E$). $\endgroup$ – Eric Wofsey Mar 15 at 22:45
  • $\begingroup$ @EricWofsey You are right thank you. $\endgroup$ – mouargmouarg Mar 15 at 22:49
  • $\begingroup$ @SantanaAfton Euh, he/she is right $\mathbb{R}^+$ is indeed closed $\endgroup$ – Thinking Mar 16 at 7:46
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You are right, it's more clear to me what's happening now. Thank you. $\endgroup$ – mouargmouarg Mar 15 at 22:56
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.