Is $\overline{\langle2\rangle\cdot(4\Bbb Z+1)}=\langle2\rangle\cdot(4\Bbb Z_2+1)$ in $\Bbb Z_2$?

$\langle2\rangle$ is the set of powers of $2$ and $\cdot$ is the straightforward dot product.

I get that $\overline{(4\Bbb Z+1)}=(4\Bbb Z_2+1)$ where $\Bbb Z_2$ is the 2-adic integers. But what about if we introduce powers of $2$ as factors? Tentatively I would say it seems the given proposition follows because $2b$ is simply the centre of the ball $b$ but then I wonder about the number $0$ which would be in one but not the other, so I'm confused!

Maybe then:

$\overline{\langle2\rangle\cdot(4\Bbb Z+1)}=\overline{\langle2\rangle}\cdot(4\Bbb Z_2+1)$ in $\Bbb Z_2$?

  • $\begingroup$ What does $\langle2\rangle$ mean here? $\endgroup$ May 7, 2019 at 11:47
  • $\begingroup$ @HenningMakholm the set of powers of 2 $\endgroup$ May 7, 2019 at 11:51
  • $\begingroup$ And the product is just each number in one set multiplied by each number in the other? No summing like for products of ideals? $\endgroup$ May 7, 2019 at 11:56
  • $\begingroup$ @HenningMakholm yes, just the straight dot product $\endgroup$ May 7, 2019 at 11:59

1 Answer 1


There's no reason to assume that the elementwise product of sets plays particularly well together with taking closure.

For example, in good old $\mathbb R$ consider the set $A=\mathbb Z\cup\{1/n\mid n\in\mathbb N_+\}$. This is a closed countable set, so $\overline A\cdot \overline A$ is countable. However $A\cdot A$ gives you $\mathbb Q$, so $\overline{A\cdot A}$ is much larger than $\overline A\cdot \overline A$.

So you shouldn't hope to compute your product just by symbolic guesswork -- if it has a nice description it will be more a matter of luck.

In your case you're right that $0$ is in $\overline{\langle 2\rangle\cdot(4\mathbb Z+1)}$, but is not in $\langle 2\rangle\cdot(4\mathbb Z_2+1)$, so those sets are definitely not equal.

It does look to me like $\overline{\langle 2\rangle\cdot(4\mathbb Z+1)}=\overline{\langle 2\rangle}\cdot(4\mathbb Z_2+1)$, but to be sure of that you need an ad-hoc proof.

  • $\begingroup$ Thanks. My intuition says the dot product multiplies and that the prime 2 is independent of $\Bbb Z_2^{\times}$, multiplication-wise, i.e. it only changes $\langle2\rangle$. In other words $\forall p:2^px:x\in\Bbb Z_2^\times$ leaves $x$ unchanged. In other words it shifts binary strings but can't change them. Therefore it can't change the $4\Bbb Z_2+1$ component of the closure. $\endgroup$ May 7, 2019 at 12:59

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