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 !

  • $\begingroup$ 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

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.

  • $\begingroup$ You are right, it's more clear to me what's happening now. Thank you. $\endgroup$ – 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.

  • $\begingroup$ 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

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