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

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 $$\Bbb Z_2$$?

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

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 $$\overline A\cdot \overline A$$.
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.
• 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. Commented May 7, 2019 at 12:59