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P-adic numbers are complete in one sense and incomplete in another sense. Is it so?

Firstly, does not complete mean connected? I read somewhere that there is not intermediate value theorem for p-adics because they are not connected. (if I am correct).

It seems I need elaboration of this "It can be shown that the rationals, together with the $p$-adic metric, do not form a complete metric space. The completion of this space can therefore be constructed, and the set of $p$-adic numbers $\mathbb{Q}_p$is defined to be this completed space."Math World.

How is the completion constructed with same metric and same numbers?

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    $\begingroup$ p-adic numbers are a stone space: totally disconnected, compact and Hausdorff; I'm not familiar with the proof of this and take it as a fact. A complete space, therefore, need not be connected. As another example of this fact you can take a disjoint union of two closed sets in $\mathbb{R}$. Lastly, I don't understand what you mean in the last question. $\endgroup$ – Andy Mar 23 '13 at 9:55
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    $\begingroup$ Just one more thing: I think that considering the algebraic construction of the p-adic numbers is better towards understanding how they behave (this is just an opinion though): the inverse limit of $\ldots \to \mathbb{Z}/(p^{n-1}) \to \mathbb{Z}/(p^n) \to \ldots$ to construct $\mathbb{Z}_p$ and then taking the field of fractions for that ring. $\endgroup$ – Andy Mar 23 '13 at 10:17
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    $\begingroup$ Complete does not mean connected. $\endgroup$ – Asaf Karagila Mar 23 '13 at 15:05
  • $\begingroup$ It’s not the “same metric” in any sense: each different $p$ defines its own different metric, all different from the standard Archimedean metric on $\mathbb Q$. $\endgroup$ – Lubin Mar 23 '13 at 15:41
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    $\begingroup$ @Andy $\mathbf Q_p$ is locally compact, not compact. $\mathbf Z_p$ is compact. So the claim that 'the p-adic numbers are a Stone space' is false. $\endgroup$ – Tomo Apr 1 '16 at 22:47
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To see that $\mathbb Q$ is incomplete under the $p$-adic valuation, it suffices to find an element of $\mathbb Q_p$ not in the rationals. For $p=2$, $\sqrt{-7}$ will do, for $p=3$, $\sqrt7$ will do, and for $p>3$, the field $\mathbb Q_p$ contains all $p-1$ roots of unity of order $p-1$. The existence of these irrationalities in the $p$-adics drops out of Hensel’s Lemma, the basic fact of $p$-adic life.

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After re-reading your question I think I understand what you want explained: take $\mathbb{Q}$, define the $p$-adic absolute value on it as $\left| x \right|_p = p^{-n}$, when $p$ is a factor in the unique factorization of $x$ ($ x = p^n a/b$, and $p$ does not divide $a$ or $b$), if $p$ is not part of the unique prime divisors as above, we say that the absolute value with respect to that is $1$.

Now we define a metric on the rationals as follows: $d(x,y) = \left| x - y \right|_p$. With respect to this metric the rationals are not complete (I can't come up with an example though) and I believe what I've read. We now take the completion of the rational with respect to this metric. The following is the general way to do it:

Let's take all Cauchy sequences with value in $X$ and let's call this set $M$. An element $x \in M$ is a sequence $x_n$; let us define $d(x,y) = \lim d(x_n.y_n)$ for all $x,y \in M$. This is not a metric, but having distance $0$ is an equivalence relation and if we take $M / \sim$, where $\sim$ is "having distance $0$", then we have a metric space with the metric defined above.

We now need to recover the points of $x$ as Cauchy sequences, in order to find $X \subseteq M/\sim$. The most logical way to do this is consider a point $x \in X$ as a sequence $x_n \in M$ converging to $x \in X$. Once we do this, it is clear that $X \subseteq M/\sim$.

If you do this for $\mathbb{Q}$ and the metric above defined, then you have $\mathbb{Q}_p$.

The question about completeness and connectedness is answered in my above comment. And yes, you're right: the intermediate value theorem relies on the connectedness of the space, thus the $p$-adic numbers do not have that property.

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    $\begingroup$ This is good, but my own preference is to think of the $p$-adic numbers as infinite $p$-ary expansions, but extending infinitely to the left instead of to the right. If there are no non-zero digits to the right of the radix point, your expansion represents an element of $\mathbb Z_p$. $\endgroup$ – Lubin Mar 23 '13 at 15:44
  • $\begingroup$ @Lubin: As I said in a comment to the question I prefer the algebraic approach of taking the field of fractions of the inverse limits of the $\mathbb{Z}/p^n$, by which you see clearly the fact you state about the roots of unity. I should also add that this might be because I never had to deal with p-adics, not once: what I know is mainly out of wikipedia entries and some thought. $\endgroup$ – Andy Mar 23 '13 at 16:02
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    $\begingroup$ Andy, of course the more abstract approach gives a deeper understanding. But I like to play around and do direct computations by hand, and the $p$-ary expansion viewpoint is superb for doing that. $\endgroup$ – Lubin Mar 24 '13 at 4:25
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Alternatively, one can verify that, e.g., the sequence $(x_n)_{n \in \mathbb{N}}$ defined by $$x_n = \sum_{j = 0}^n{p^{j^2}}$$ is Cauchy with respect to the $p$-adic valuation, but does not have a limit in $\mathbb{Q}$.

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  • $\begingroup$ I wonder if this is somehow related to the example given at the end of this answer or this thread. $\endgroup$ – Watson Feb 13 '17 at 17:26

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