Set of sequences -roots of unity

Question

Consider $G_n$ as the multiplicative cyclic group given by the $n^{th}$ roots of unity.

$$G_n = \left\{ e^{ 2ik\pi/n} \mid 1\leq k \leq n \right\}$$

Now construct a sequence from each $G_n$ by $(e^{2i1\pi/n}, e^{2i2\pi/n},\dots, e^{2ik\pi/n},\dots,e^{2in\pi/n}, 0, 0, 0,\dots$).

Now consider the set S of all such sequences, $$S=\left\{(e^{2i1\pi/n}, e^{2i2\pi/n},\dots, e^{2ik\pi/n},\dots,e^{2in\pi/n}, 0, 0, 0,\dots) : 1\leq n \lt \infty\right\}$$ Will I be right in concluding that no matter what operation I impose on $S$, the structure of $S$ will never be a semi-group/vector space/normed vector space?

I wondered about the practical utility of creating $S$. Any help?

An observation is that $(1,0,0,\dots)$ is the only limit (obtained by pointwise convergence of first sequence) contained by $S$ (when $S$ is viewed as a subset of $C^n$. It contains no other elements that are obtained by pointwise convergence of each sequence. The pointwise limit of each sequence is of the form $(1,1,....,\underbrace{1}_{n^\text{th}\text{ place}},0,0,\dots)$.

Answer 1

Let us denote by $s_n$ the sequence constructed from $G_n$. Let us define $$s_m*s_n=s_d{\rm\ where\ }d={\rm lcm}(m,n)$$ Voila! $S$ is now a semi-group!

Answer 2

Will I be right in concluding that no matter what operation I impose on $S$, the structure of $S$ will never be a semi-group/vector space/normed vector space?

Adam Hughes has essentially answered the question in the comments. I am simply going to pile (a lot) more detail onto his explanations. In my opinion, the most honest answer to this question is:

Technically no, but yes.

To expand on this briefly before getting heavy: If someone were to come up and ask me this question, I would simply say, "yes, that is correct". This is (formally) a complete lie, and if I was asked to explain logically why that was correct, I would be completely unable to, because it's not true. This is a reasonable follow-up question, and I am much more likely to be caught in this lie than in saying some other blatantly false thing like "$.999...\neq 1$", so you might wonder what would drive me to this kind of behavior. That is what I will try to get at with the rest of this answer.

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Since I will spend a lot of time discussing technicalities and intuitions, here are a few that I just want to smooth over right away:

• $S$ is not a subset of $\Bbb C^n$ for any $n$; in fact, no element of $S$ is an element of $\Bbb C^n$ for any $n$.

• Elements of $S$ are probably not usefully thought of as sequences, although they are technically sequences.

• Instead, we start with a set called $\Bbb C^\Bbb N$ which you can think of as containing complex vectors with (countably) infinitely many elements.

• It is easy to check that $\Bbb C^\Bbb N$ is a vector space.

• But notice that $$\{(1,0,0,\dots); (0,1,0,\dots); (0,0,1,\dots), \cdots\}$$ is not a basis of $\Bbb C^\Bbb N$. The span of this set is called $\Bbb C^\infty$, and it is easy to check that $S\subset\Bbb C^\infty$.

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To explain why the statement is not true, a counterexample will suffice: I will prove that $S$ is a field; in fact, it is isomorphic to $\mathbb Q$. Fortunately, the counterexample will also elucidate why my answer is "yes".

We begin by noting that $S$ is countable, and therefore there exists a bijection $S\to\mathbb Z^+$. We can do better than that, we can write one down explicitly. Define the support of a vector in $\Bbb C^\infty$ to be the number of nonzero entries: this number is finite, non-negative, and zero only for the zero vector, so the support $\sigma$ is a well-defined function from $\Bbb C^\infty\smallsetminus\{0\}$ to $\mathbb Z^+$. It follows from the definition on $S$ that the restriction of $\sigma$ to $S$ is a bijection.

Let $\phi$ be your favorite bijection $\Bbb Z\to \Bbb Q$. Remember that all bijections have inverse functions. Then define $\oplus$ and $\otimes$ as functions $S\times S\to S$ such that

$$s\oplus t = \sigma^{-1}\Big(\phi^{-1}\Big( \phi(\sigma(s))+\phi(\sigma(t)) \Big)\Big)$$ $$s\otimes t = \sigma^{-1}\Big(\phi^{-1}\Big( \phi(\sigma(s))\cdot\phi(\sigma(t)) \Big)\Big)$$

The inner addition and multiplication are the ordinary operations on $\mathbb Q$. These functions are well-defined, and the field axioms follow by the injectivity of $\sigma^{-1}$ and $\phi^{-1}$. For the incredulous, the associativity of multiplication is proved below; others are similar.

$$s\otimes (t\otimes u) = \sigma^{-1}\Bigg(\phi^{-1}\Bigg( \phi(\sigma(s))\cdot\phi\Big(\sigma\Big( \sigma^{-1}\big(\phi^{-1}\big( \phi(\sigma(t))\cdot\phi(\sigma(u)) \big)\big)\Big)\Big) \Bigg)\Bigg)$$ $$= \sigma^{-1}\Bigg(\phi^{-1}\Bigg( \phi(\sigma(s))\cdot\Big( \phi(\sigma(t))\cdot\phi(\sigma(u)) \Big) \Bigg)\Bigg)$$ $$= \sigma^{-1}\Bigg(\phi^{-1}\Bigg( \Big(\phi(\sigma(s))\cdot \phi(\sigma(t))\Big)\cdot\phi(\sigma(u)) \Bigg)\Bigg)$$ $$= \sigma^{-1}\Bigg(\phi^{-1}\Bigg( \phi\Big(\sigma\Big( \sigma^{-1}\big(\phi^{-1}\big( \phi(\sigma(s))\cdot\phi(\sigma(t)) \big)\big)\Big)\Big) \cdot \phi(\sigma(u)) \Bigg)\Bigg) = (s\otimes t)\otimes u$$

[It's the sort of proof that only an algebraist could ever love.]

Once we are convinced that $S$ is a field under these two operations, it is easily determined that $\phi\circ\sigma$ is a homomorphism $S\to\Bbb Q$; since it is bijective, we have that $S\cong \Bbb Q$.

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My hope is that, after seeing this construction, you automatically understand why my answer is "yes". If this is true, you can skip this section.

Else, let us consider this famous quote of Halmos (emphasis mine):

Don't just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?

If the theorem is "$(S,\oplus, \otimes) \cong (\Bbb Q,+,\cdot)$ as fields", and the proof is as above, one might wonder about these two instructions. What is the hypothesis? I posit that the hypothesis is many-layered: in order of depth, first it is the construction of $\oplus$ and $\otimes$, then the construction of $S$, and then the construction of $(\Bbb Q, +, \cdot)$.

The construction of $\oplus$ and $\otimes$ is clearly of critical importance, and the proof uses them many, many times to prove the field axioms and then the isomorphism.

But what about the construction of $S$? Go look at it. I'll wait.

Did you see it?

The proof never uses the construction of $S$. It says it does, but if you read carefully, you see that the only fact that it uses about $S$ is that $\sigma$ is a bijection $S\to\Bbb Z^+$; that is, that $S$ is countable.

We are then forced to conclude the following theorem, because the proof is identical:

For a suitable construction of $\oplus$ and $\otimes$, any countable set $A$ has $(A,\oplus, \otimes) \cong (\Bbb Q,+,\cdot)$ as fields.

Well, that was unexpected.

How about the construction of $\Bbb Q$? Go back and look again.

Really, do it.

Your eyes are not deceiving you. The construction of $\mathbb Q$ is never used. Again, it says it is, but the only facts that we are using about $\mathbb Q$ is that it is countable (for the existence of $\phi$) and that it is a field (for the well-definition of $\oplus$ and $\otimes$).

That means we are forced to conclude this still more general theorem:

For any countable field $F$ and a suitable construction of $\oplus$ and $\otimes$, any countable set $A$ has $(A,\oplus, \otimes) \cong F$ as fields.

But wait, it gets better.

The only reason countable shows up is because of the intermediate $\mathbb Z^+$. If we had a single bijection $A\to F$, it could play the role of $\phi\circ\sigma$, and the proof goes through. Therefore, we can strike countable from the theorem above, replacing with the additional hypothesis that $|F|=|A|$.

And now the coup de grâce: where do we use the fact that $F$ is a field? The answer is, to prove the field axioms for $A$. But the technique used to prove them can be used to do this for any algebraic structure. If $F$ is a ring, then $A$ is a ring. If $F$ is a quasigroup, then $A$ is a quasigroup. And so on. Now we have a family of nearly-identical theorems about arbitrary sets being algebraic structures.

But for this generality, we pay a very, very high price. As Adam so eloquently put it (emphasis mine):

There's nothing of the original set's personality left with [this] construction.

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Okay, so… what, exactly? These are pathological examples, clearly. We just have to exclude them, and then we can focus on the real ones.

Except it's not that easy. The problem is, every isomorphism between two algebraic structures is by definition a bijection, and is therefore a candidate for $\phi\circ\sigma$. Therefore, if we exclude fields made by this construction, we exclude every field. Even good, wholesome fields like $\Bbb R$ and $\Bbb Z_{81}$.

This is a sad state of affairs. What are we to do?

The unfortunate answer is that we cannot do anything. All sets are all fields (modulo cardinality concerns), all groups are rings, all monoids are $\Bbb C$-algebras; we must accept it. The only thing to do is admit that most fields are !@#%^&!@ and try to focus on the ones that aren't.

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So let us interpret your question with this in mind. Is there an algebraic structure on $S$ that isn't !@#%^&!@ ? The important thing to remember is that !@#%^&!@ is not formally defined, so the answer to this question is inherently subjective. Moreover, a negative answer is almost never possible. There may be a theorem that strongly suggests the answer is no, but you cannot actually be certain that one of the pairs of operations skirts around the hypotheses of that theorem and yet is less !@#%^&!@ than you would have guessed.

However, there is one way that we can safely say that a group is not !@#%^&!@ with very little controversy. There is some group that is not !@#%^&!@, for example $\Bbb C$. If we can represent $S$ as a subgroup (or at least isomorphic to a subgroup) of this group, then it is also not !@#%^&!@. In other words, not-!@#%^&!@-ness is a hereditary property.

Along these lines, here is some evidence that $S$ does not have this property:

• $S$ is not a subspace of $\Bbb C^\infty$.

• Every element of $S$ lies in the infinite direct product of unit circles union the origin. But $S$ is not a submonoid of this monoid.

• $S$ is not a submagma of any countable product of finite magmas, despite being defined by countably many finite magmas (indeed, groups).