# Generalizing limits of sums, products, and quotients of sequences to abstract topological spaces?

Introductory real analysis books usually prove a list of properties about limits of sequences of real and complex numbers. Suppose $\lim x_n=x$ and $\lim y_n=y$. Then:

1. $\lim (x_n+y_n) = x+y$
2. $\lim c x_n = cx$ (where $c \in F$ for some field $F$)
3. $\lim x_n y_n = xy$
4. $\lim x_n^p=x^p$
5. $\lim \frac{1}{x_n} = \frac{1}{x}$ (provided for all $n$, $x_n \ne 0$ and $x \ne 0$)
6. $\lim \frac{x_n}{y_n}=\frac{x}{y}$ (provided for all $n$, $x_n \ne 0$, $y_n \ne 0$, $x \ne 0$ and $y \ne 0$)

However, these results can be generalized beyond sequences of real & complex numbers to more abstract sequences in more general metric and topological spaces. I would like to prove these results for an arbitrary metric space. However, the problem is, as far as I know, these operations (addition, multiplication and division) are only defined for numbers. So in an abstract metric or topological space, does $x_n+y_n$ have meaning?

How would one generalize these properties to abstract metric and topological spaces? Should you start with topological field extensions, topological algebras over a topological field, normed algebras, or something else? And just how general can you make these results? What is the most general topological space in which the operations $x_n+y_n$, $x_n \times y_n$, $\frac{x_n}{y_n}$, etc. make sense and all 6 properties of limits hold? Thanks for your thoughts!

• Please stop trying to bump this question by making trivial edits. It's really annoying. – user98602 Sep 21 '15 at 4:48
• Please stop bumping this question by making trivial edits. It's really annoying. – Pedro Tamaroff Sep 21 '15 at 4:52

The group of real numbers $(\Bbb R,+,0,-)$ with its usual topology is a topological group: The addition $+:\Bbb R\times\Bbb R \to \Bbb R$ (where $\Bbb R\times \Bbb R$ has the product topology) and the negation $-:\Bbb R \to \Bbb R$ are continuous maps. These properties of $\Bbb R$ are exactly what leads to formulas like $$\lim(x_n+y_n) = x+y\quad \text{ and } \quad \lim(-x_n)=-x$$ the reason being that the sequence $(x_n,y_n)_n\subseteq\Bbb R\times\Bbb R$ converges to $(x,y)$, thus, as a continuous map preserves limits, the image sequence $(+(x_n,y_n))_n = (x_n+y_n)_n$ converges to the sum $+(x,y) = x+y$. By the same reasoning, we have $(-(x_n))_n\to -x$.

Replacing addition $+$ with multiplication $\cdot:\Bbb R\times\Bbb R\to\Bbb R$, though, we don't get a group but rather a monoid $(\Bbb R,\cdot,1)$, and it is topological in the sense that $\cdot$ is continuous. By the same reasoning as above, we have $\lim(x_ny_n)=xy$.
We regain a topological group, however, if we restrict to non-zero real numbers $\Bbb R_\times$, since the map $x\mapsto 1/x$ is continuous. So it follows that $\lim(1/x_n)=1/x$.

Examples are topological groups are

• The circle $S^1$ with the multiplication and topology inherited from the complex numbers.
• The group $\text{Aut}(X)$ of homeomorphisms on a compact Hausdorff space $X$. The multiplication is the composition of two maps, and the topology is the compact-open topology
• Any normed vector space $(V,||-||)$ has a topology induced by the norm, and this topology makes the addition $+$ and the negation $-$, as well as scalar multiplication $\Bbb R\times V\to V$ continuous. That's why we speak of topological vector space.

I'm going to address your desire for a structure where all these properties hold: We assume that our structure $X$ has an addition, an inner multiplication, and an outer scalar multiplication with a field $F$, subject to the following rules

1. $(x+y)+z = x+(y+z)$
2. $x+y = y+x$
3. There is an element $0$ such that $0+x = x+0 = x$
4. For each $x\in X$, there is a $-x\in X$ such that $x+(-x) = 0$.
5. $(λ\mu)x = λ(\mu x)$
6. $\lambda(x+y) = λx + λy$
7. $(λ+\mu)x = λx + \mu x$
8. $(xy)z = x(yz)$
9. $xy = yx$
10. There is an element $1_X\in X$ such that $1_X x = x1_X = x$
11. For each $x\in X\setminus\{0\}$, there is an $x^{-1}=\frac1x$ such that $x/x=1_X$
12. $(x+y)z = xz+yz$
13. $(λx)y = λ(xy)$

where $x,y,z\in X$ and $λ,\mu\in F$, and $x/y$ is another notation for $xy^{-1}=y^{-1}x$

One could subsume these rules by saying that $(X,+,0,-,\cdot,1_X,^{-1})$ is a field satisfying the rules 5, 6, 7, and 13. Note that we can deduce some more properties from these

• $0_F x = λ0_X = 0_X$
• $-(λx) = (-λ)x = λ(-x)$
• $1_F x = x$. For if $1_Fx=y$, then $$1_F x = (1_F1_F)x = 1_F(1_F x) = 1_F y$$ thus $1_F(x-y) = 0$. Now if $x\ne y$, then $z=(x-y)$ has an inverse, hence $$1_F1_X = 1_F(zz^{-1}) = (1_Fz)z^{-1} = 0_Xz^{-1} = 0_X$$ which again implies that $$λw = (λ1_F)(1_Xw) = (λ(1_F1_X))w = λ0_Xw = 0_Xw = 0$$ for any $λ\in F$ and $w\in X$, which is probably not what you want.
• The map $i:λ\mapsto λ1_X$ is a field homomorphism. For if $λ,\mu\in F$, then $(λ+\mu)1_X = λ1_X + \mu1_X$ and $(λ\mu)1_X = λ(1_X(\mu1_X)) = (λ1_X)(\mu1_X)$. So either $i\equiv0$, or $i$ is an embedding of $F$ into $X$. But if $i\equiv0$, then $λx=λ(1_Xx)=(λ1_X)x=0_Fx = 0$ for any $λ\in F$ and $x\in X$.

The last item is why we can regard $F$ as a subfield of $X$, so whacka's remark about $X/F$ being a field extension is justified.

Now $X$ is equipped with a topology, otherwise we could not talk about limits of sequences. Note that if we equip $X\times X$ with the product topology, then sequences $(x_n)_n$ and $(y_n)_n$ converge to $x$ and $y$, respectively, if and only if $(x_n,y_n)_n$ converges to $(x,y)$ in $X\times X$. So the property $\lim_{n\to\infty}(x_n+y_n)_n = x+y$ simply means that addition $+:X\times X\to X$ preserves limits of convergent sequences, thus is sequentially continuous (as opposed to being continuous, which is slightly stronger). So the limit properties you stated hold if and only if addition, multiplication, and inversion are sequentially continuous. One could call such a structure a sequentially topological field or something like that.

• But these results aren't specific to $\mathbb{R}$. Indeed, they hold in $\mathbb{C}$ and more abstract metric spaces. That's the point of my question - what's the most abstract / general we can make this result? – Mathemanic Aug 30 '15 at 18:29
• @EthanAlvaree: I would say, the most general objects where these results hold are objects that are equipped with a topology (because of $\lim$) as well as a binary relation such that these two structures are compatible with each other, and that's just a topological monoid (top. group, top. vector space). – Stefan Hamcke Aug 30 '15 at 18:35
• Thanks for your reply and your mention of vector spaces. However, take the property that $\lim x_n y_n = xy$. This property doesn't hold in a vector space, because multiplication of vectors is undefined. So I am thinking the most abstract we can have these properties is for a ring or field. But I am not sure if there might be more abstract spaces where these results still hold. – Mathemanic Aug 30 '15 at 18:40
• @EthanAlvaree: Oh, I see, you want all these formulas to be defined and to hold at once, in a single structure. Well, then you would have topological field, i.e. a field $(F,+,0,-,*,1,^{-1})$ such that $(F,+,0,-)$ and $(F\setminus\{0\},*,1,^{-1})$ are topological groups. – Stefan Hamcke Aug 30 '15 at 18:43
• @EthanAlvaree: Actually, since you don't require $*$ to be commutative, a skew field would suffice. And since you want to multiply $x_n$ by a scalar $c\in\Bbb R$, this skew field would at the same time have to be a topological vector space. That means it is a topological algebra, but one in which you can divide elements. – Stefan Hamcke Aug 30 '15 at 18:48

Indeed the algebra of limits holds (for sums and scalar multiples) in any normed space. The proof is identical. The remaining algebra of limits hold in any normed algebra. Note here that the norm on an algebra imust satisfy $\lVert xy\rVert\leq\lVert x\rVert\lVert y\rVert$ for all $x,y$ in the algebra. Also note that non-zero must be replaced by invertible in the multiplication of the algebra.

A particular example of such an algebra is the space of all continuous, complex-valued functions on a compact Hausdorff space, equipped with the supremum norm and point-wise operations.

(n.b. I typed this on my phone, so it is undoubtedly filled with the wrong words and bad typos)

• Hi Sam, thanks for your reply. What do you mean by a normed "algebra"? I have studied abstract algebraic structures such the ring, field, vector space, etc. But is an "algebra" in the same category as these? And if so, how would you define it? – Mathemanic Aug 30 '15 at 18:34
• @EthanAlvaree en.wikipedia.org/wiki/Algebra_over_a_field – oldrinb Aug 30 '15 at 18:40
• @SamM, do these properties hold in any normed algebra, or do we also require it to be "topologically complete" (en.wikipedia.org/wiki/Complete_metric_space) or "completely metrizable" (en.wikipedia.org/wiki/Completely_metrizable_space)? – Mathemanic Sep 11 '15 at 20:01
• It isn't a property of completeness. What's really going on here is that the operations $(x,y)\mapsto x+y$ and $(x,y)\mapsto xy$ are continuous with respect to the topology on the space (see the comments in the above). In other words, whenever we have a set $X$, which has some kind of algebraic structure and a topological structure for which the algebraic operations are (jointly) continuous. Then the identities you want. – SamM Sep 11 '15 at 20:06

To talk about things like addition, multiplication, multiplicative identity, unlimited division, scalars from $F$ and for the notation $\frac{a}{b}$ to make sense, you're already talking about a field extension of $F$.

Then (almost by definition) a structure satisfies your list of properties $\iff$ it's a topological field which is also a field extension of $F$. There is some redundancy in the list:

• $(3)\implies(2)$
• $(3)\implies (4)_{p\ge0}$
• $(3)\land(5)\implies(4)_{p\in\Bbb Z}$
• $(4)_{p\in\Bbb Z}\implies(5)$
• $(3)\land(5)\implies(6)$

Just throw away $(4)$ from the outset. We can get some other structures by discarding other conditions:

• If one throws out $(2)$, then we have a topological field with no relation to $F$. If it is locally compact and not discrete, this is called a local field, and the topology is induced from a metric which is itself induced from an absolute value function.

• If one throws out $(6)$ then multiplication isn't necessarily commutative so it is a topological division algebra, aka skew field, that contains the field $F$.

• If one throws out $(2)\land(6)$, it's a topological skew field with no relation to $F$.

• If one throws out $(5)\land(6)$ then it is a topological ring that contains $F$.

• If one throws out $(2)\land(5)\land(6)$ then it is a topological ring.

• If one throws out $(2)\land(3)\land(5)\land(6)$ then it's a topological abelian (additive) group.

• If one throws out $(1)\land(2)\land(5)\land(6)$ then it's a topological (multiplicative) group.

• If one throws out $(3)\land(5)\land(6)$ then it's a topological vector space over $F$.

Essentially, first figure out the algebraic structure you want, then add the word topological in front of it. One can still get things that don't have names, I suppose. (For instance, consider a topological ring in which inversion is continuous on the subspace of two-sided units, or take one of the above listed objects and append topological conditions on.)

• Thank you so much for your answer. You said a structure (almost by definition) satisfies these properties if and only if it is a topological field extension of $F$. However, the other answers & comments talk about a topological algebra (vector space with multiplication) over a field $F$. I am not sure which structure to use for these limit rules, but I'm currently leaning toward topological algebra. Is one space more general than the other? – Mathemanic Sep 18 '15 at 2:20
• Thinking about it a little bit more, I realize that either way, I would need the notion of a topological field. Because even topological algebras are defined over topological fields. The confusing thing is, the topological field would satisfy the limit properties AND the topological algebra over that topological field would also satisfy the same limit properties, simultaneously. Am I understanding correctly? – Mathemanic Sep 18 '15 at 2:23
• @EthanAlvaree Your condition (5) says every nonzero element has an inverse, and condition (6) implicitly assumes multiplication is commutative, so together they imply the structure is itself a field. Any field that contains $F$ is called a field extension of $F$. Most often one uses the term "algebra" to refer to a ring that contains a given field. Also, sure, I see nothing confusing about both a field and a vector space over it being topological. – whacka Sep 18 '15 at 2:31
• I am just learning that an algebra (over a field) is a vector space endowed with multiplication $\cdot: V \times V \to V$, although I have never seen multiplication of vectors defined except for the dot product, which doesn't satisfy our property #3. en.wikipedia.org/wiki/Algebra_over_a_field – Mathemanic Sep 18 '15 at 2:52