1

enter image description here

There are two things I am confused about

1) In the second line of the proof, How does $a=0$ imply that $G$ contains the $\epsilon-$dense set?

2) In the last line of the proof, How does the definition of $a$ imply that $r=0$?

gbd
  • 1,963
  • The conclusion that $G$ contains $\Bbb Z \epsilon$ seems too quick. What can be said, in the case that $a = 0$, is that $G$ contains arbitrary small $\epsilon > 0$ (i.e., for every $\delta > 0$, there is a $0 < \epsilon < \delta$ with $\epsilon \in G$). That's sufficient to conclude that $G$ is dense. – Magdiragdag May 03 '17 at 20:00
  • @Magdiragdag, Why does $G$ being dense imply that $\mathbb{R}=G$? – gbd May 03 '17 at 20:02
  • Is that a third thing you're confused about? It is because $G$ is closed. – Magdiragdag May 03 '17 at 20:05
  • @Magdiragdag, I just don't think I clearly understand the full proof. Especially the first part. – gbd May 03 '17 at 20:10
  • I think it is worthwhile to consider where this proof breaks down if $G$ is not closed, for instance if $G = \Bbb Q$. – Magdiragdag May 03 '17 at 20:14

2 Answers2

1

Both questions are answered by the definition of $a$ as it happens!

For question 1, if $a = 0$ then because we define $a$ as the infimum of the set $\{t\in G | t >0\}$ we must have that there are elements of $G$ arbitrarily close to 0 (as Magdiragdag points out, not every positive $t$ need be in the group), and we can write $t$ as $\epsilon$ to make clear that we're talking about very small values of $t$. All integer multiples of an element of $G$ are contained in the group, so $\mathbb{Z}\epsilon$ is contained in the group. This means we have a countable (since $\mathbb{Z}$ is countable) set of elements of $G$ arbitrarily close to any other element of $G$, i.e. a dense subset of $G$. Then $G$ being closed yields the equality with $\mathbb{R}$.

For question 2, by writing $r=s-pa$ we see that $r\in G$ (because $s$ and $pa$ are both elements of $G$), and we have stated that $0\leq r<a$. However, since $a$ is the infimum of non-zero elements of $G$ by definition the only element of $G$ smaller than $a$ is $0$. So $r=0$.

postmortes
  • 6,338
  • 1
    It is not automatically true that any $t > 0$ is an element of $G$ if $a = 0$. What is true is that $G$ must contain elements arbitrarily closed to $0$. (For a counterexample, consider $G = \Bbb Q$. This is a (non-closed) subgroup of $\Bbb R$ for which $a = 0$ and of course it does not contain all $t > 0$). – Magdiragdag May 03 '17 at 20:10
  • @Magdiragdag -- thank-you, well spotted! I've updated the proof accordingly and credited you. – postmortes May 03 '17 at 20:16
1

1) If $a=0$ then for every $\epsilon>0$ there is $x\in G\cap (0,\epsilon)$. $G$ is group, thereby $\mathbb{Z}x\subset G$, and $\mathbb{Z}x$ is $\epsilon$-dense by definition.

2) Assume, that $r>0$. But $r\in \{t\in G: t>0\}$, and $r<a=\inf\{t\in G: t>0\}$. Contradiction.

Przemek
  • 855