# Example of different topologies with same convergent sequences

It's well known that for metric spaces the following is true

Let $X$ be a space with two different metrics $d_1,d_2$ such that the two topological spaces $(X,d_1),(X,d_2)$ have the same convergent sequences. Then the two topologies are the same.

Now looking at a space $X$ with two topologies $\tau_1,\tau_2$ this is not true any more, i.e. if the topological spaces $(X,\tau_1), (X,\tau_2)$ have the same convergent sequences the topologies may differ!

A classical example is $l^1$ due to Issai Schur.

So my questions are:

1. Is there a more topological/simpler example?
2. I just know the "standard" functional analysis proof of Schur's lemma. On Wikipedia they refer to his article in "Journal für die reine und angewandte Mathematik, 151 (1921) pp. 79-111". I didn't work through the paper (and I won't) but just skimming through the paper, I don't see how we can deduce from the paper, that $l^1$ has the Schur property. If there is someone who's familiar with the paper a short argument would be appreciated.

I'm not sure if I should place the second question here. If not, let me know and I will post a new one.

Thx and cheers

math

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Just a general remark: why don't you give it some time before accepting an answer? There's no hurry to accept answers, and I believe this question would be one that deserves to be open for a bit --- there are quite a few topologists here who could contribute more and maybe even nicer examples. Also, only part 1 was addressed so far, so accepting an answer will discourage people from reading the question properly and add things of substance. –  t.b. Oct 28 '11 at 15:44
@t.b.: Actually I noticed to late JHD's answer! Sorry for that. About question 2: I really just skimmed through it, so if the answer is that easy, let me know. I don't want to waste someone's time –  math Oct 28 '11 at 15:48
About your second comment: I guess that my second question could seem to be impolite, since it looks like that I'm too lazy to bring up the effort. –  math Oct 28 '11 at 15:54
Well, as I said, I read the paper quite some time ago, and I'll make the connection to Schur's property somewhat more explicit. It is contained in his "Hauptsatz" but it needs some working out. Check back in a few hours! And... nothing impolite about asking that question! –  t.b. Oct 28 '11 at 15:57
Okay, as promised I updated my answer in order to address question 2. –  t.b. Oct 28 '11 at 19:10

Edit: Since question 1 was beautifully answered by JDH, I'll leave it at the simplest and silliest example that came to my mind — more substance (I hope) is in the second part of the answer:

On an uncountable set, both in the discrete topology and the cocountable topology (the topology consisting of the empty set and the sets with countable complement), the only convergent sequences are the eventually constant ones. Since the set is not countable, the topologies are distinct (countable sets are open in the discrete topology while they aren't in the cocountable topology).

As for question 2, the paper in question is:

J. Schur, Über lineare Transformationen in der Theorie der unendlichen Reihen, Journal für die reine und angewandte Mathematik (Crelle's Journal), 151 (1921), 79–111 (full text is behind a pay wall).

In modern terms, Schur starts out by identifying $\ell^1$ as the dual space of $c$, the space of convergent sequences via the pairing $\langle \mathbf{a},\mathbf{x}\rangle_{\ell^1, c} = \sum_{n=1}^{\infty} a_n x_n$:

Auxiliary Theorem. For a real or complex sequence $(a_n)_{n \in \mathbb{N}}$ the sum

$$a_1x_1 + a_2x_2 + \cdots$$

converges for each convergent sequence $(x_n)_{n=1}^{\infty}$ if and only if the series

$$a_1 + a_2 + \cdots$$

converges absolutely.

The proof is a rather immediate consequence of Abel's summation criterion.

Consider a sequence $(\mathbf{a}_{n})_{n =1}^{\infty} \subset \ell^{1}$, where $\mathbf{a}_{n} = (a_{n1}, a_{n2}, \ldots)$. Let $c$ be the space of convergent sequences. Note that for each sequence $(\mathbf{a}_n)_{n=1}^{\infty}$, we get a linear map $A: c \to \mathbb{R}^{\mathbb{N}}$ given by $$A\mathbf{x} = (\langle \mathbf{a}_1, \mathbf{x}\rangle, \langle \mathbf{a}_2,\mathbf{x}\rangle, \ldots).$$ Schur seeks to determine necessary and sufficient conditions to ensure that $A$ defines a linear map $c \to c$ (he calls such maps convergence-preserving: they map convergent sequences to convergent sequences).

The necessary and sufficient conditions are (part I of the Hauptsatz on page 82):

The sequence $(\mathbf{a}_{n})_{n=1}^{\infty}$ defines a linear map $A: c \to c$ if and only if

1. For every $k$ the limit $a_{k} = \lim\limits_{n\to\infty} a_{nk}$ exists.
2. Put $\sigma_{n} = \sum_{k=1}^\infty a_{nk}$ then $\sigma_n$ converges to some $\sigma \in \mathbb{R}$.
3. There exists aconstant $C$ such that for all $n$ we have $\|\mathbf{a}_{n}\|_1 \leq C$.

Moreover, if these conditions are satisfied then $\mathbf{a} = (a_k)_{k=1}^{\infty} \in \ell^1$ and putting $a = \sum_{k=1}^{\infty} a_k$ we have $$\lim_{n \to \infty} \langle \mathbf a_n, \mathbf x\rangle = (a-\sigma)\lim \mathbf{x} + \langle \mathbf a, \mathbf x\rangle,$$ in particular $\mathbf a_n \to \mathbf a$ in the weak$^{\ast}$-topology, if we identify $\ell^1 = (c_{0})^\ast$.

The second question Schur asks is: when does a sequence $(\mathbf{a}_n)_{n=1}^{\infty} \subset \ell^1$ induce an operator $\ell^{\infty} \to c$? The formula is again $A: \ell^{\infty} \to \mathbb{R}^{\mathbb{N}}$ and $$A\mathbf{x} = (\langle \mathbf{a}_1, \mathbf{x}\rangle, \langle \mathbf{a}_2,\mathbf{x}\rangle, \ldots)$$ for $\mathbf{x} \in \ell^{\infty}$ (he calls such operators $A$ convergence-generating). Since $A$ as above induces in particular an operator $A: c \to c$, the above conditions must be satisfied, so the condition must certainly be stronger. Indeed, part III of the Hauptsatz on page 82 reads:

In the above notation $A$ defines a linear map $\ell^{\infty} \to c$ (in particular the sequence $\mathbf a_n$ converges weakly to $\mathbf a$) if and only if for every $\varepsilon > 0$ there exists $l$ such that $\sum_{k=l+1}^{\infty} |a_{nk}| \lt \varepsilon$.

In the course of the proof he establishes that $\|\mathbf{a}_n - \mathbf{a}\|_{1} \;\xrightarrow{n\to\infty} \; 0$, so he shows that weak convergence implies norm convergence, as desired. To prove this, he proceeds by contradiction (see §4, p.89f). One easily reduces to the case that $\mathbf{a} = 0$ and, assuming that $\|\mathbf a_{n}\|_{1} \not\to 0$, he builds a bounded sequence $\mathbf{x}$ for which the sequence $A\mathbf{x} = (\langle \mathbf{a}_1, \mathbf{x}\rangle, \langle \mathbf{a}_2,\mathbf{x}\rangle, \ldots)$ is not convergent, contradicting weak convergence.

For the convenience of the readers, here is the Hauptsatz in full:

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Wow! Thx t.b. that's much more than I expected! Just skimming through, the connection to today's notation was (for me) not that obvious. I guess this is a common problem reading historical papers. Anyway thx again for your answer. –  math Oct 29 '11 at 18:36
@math: You're very welcome! I've slightly updated the answer. –  t.b. Oct 30 '11 at 7:12
@t.b. Very nice answer. I was wondering about some details. At one place you write that "Schur starts out by identifying $\ell^1$ as the dual space of $c$" and later you use part I of the Hauptsatz to deduce that $\mathbf{a_n}\rightarrow\mathbf{a}$ in the weak$^*$-topology of $\ell^1=(c_0)^*$. So apparantly we have both $$c^*=\ell^1=(c_0)^*,$$ and of course $c=c_0\oplus\mathbb{R}$. I take it $c$ and $c_0$ are taken as (closed) subspaces of $\ell^{\infty}$. Now I see what's bothering me : the left hand side should be an inclusion $\ell^1\subset c^*$ (maybe as a codimension one closed subspace). –  Olivier Bégassat Oct 30 '11 at 7:49
@Olivier: Thanks! I tried to be faithful to Schur's text, so I couldn't write "identifying $\ell^{1} = (c_0)^{\ast}$". Also, he doesn't really show that we have this duality (duality theory for Banach spaces came only 5-10 years later in Banach's work!), but all the ingredients are there. I guess the natural pairing between $\ell^1$ and $c$ would rather be $$\langle \mathbf a,\mathbf x \rangle = \left(\sum a_n\right)\lim \mathbf x + \sum a_n (x_n-\lim \mathbf x),$$ so yes, I agree that would probably be nicer (and it bothered me, too!). I'll have to think how to implement this in my answer. –  t.b. Oct 30 '11 at 8:02
Perhaps it's worth mentioning that in the study of matrix methods in summability theory the infinite matrices fulfilling the above conditions are sometimes useful and they are called conservative matrices. See Theorem 2.3.7 in the book Boos: Classical and modern methods in summability. Boos calls these conditions (Zn) Zeilennormenbedingung, (Sp) Spaltenbedingung, (Zs) Zeilensummenbedingung. –  Martin Sleziak Oct 30 '11 at 8:56

Here is one quick example. Let $X$ be any uncountable set, and let $\tau_1$ be the discrete topology on $X$, and let $\tau_2$ be the topology induced by the co-countable sets, that is, the complements of countable sets are open. These two topologies are not the same, but for each of them, the only convergent sequences are the eventually constant sequences. This is because every countable set is closed for each of the topologies.

Here is another example, where both spaces are Hausdorff. Let $\tau_1$ be the usual order topology on $\omega_1+1$, where $\omega_1$ is the first uncountable ordinal and the $+1$ means that we have placed a point at the top, which makes this a compact Hausdorff space. Let $\tau_2$ be the topology on $\omega_1+1$, where the top point is isolated. This space remains Hausdorff, but no longer compact. Meanwhile, however, the two spaces have exactly the same convergent sequences, since these are simply the eventually constant sequences plus the sequences that eventually stay below $\omega_1$ and converge there. The relevant fact is that every countable set of ordinals is bounded below $\omega_1$, and hence does not interact with the place where we have changed the topology.

A similar example can be made from the long line, by making a top point a limit point of what is below or by making it isolated.

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The point with the long line example is that by looking only at sequences, you cannot tell if the additional point at the top is isolated or not. –  JDH Oct 28 '11 at 15:42
I think in the second example it is referring to $\omega$, the first countable ordinal number. In fact, for $\omega_1$, we don't have any sequence of numbers below $\omega_1$ which converge to it. $\omega_1 +1$ is not compact either. –  Xiaochuan Oct 28 '11 at 18:13
@Xiaochuan: In the order topology successor ordinals are compact. This is because if you cover the whole space you can find a decreasing chain of ordinals which is finite, and each ordinal marks a set in the cover. Thus finding a finite subcover. –  Asaf Karagila Oct 28 '11 at 18:23
@Xiaochuan, I meant $\omega_1$, not $\omega$, and to my way of thinking, the fact that no countable sequence converges to $\omega_1$ from below is precisely what makes the example work. And as Asaf says, $\omega_1+1$ is compact, which can be proved by transfinite induction. –  JDH Oct 29 '11 at 1:41
@JDH, I didn't understand. Got it, thank you. –  Xiaochuan Oct 30 '11 at 18:04

Many spaces whose topologies aren’t determined by their convergent sequences (i.e., non-Fréchet spaces) will serve in the same way as JDH’s examples. In particular, any space with a non-isolated point that is not the limit of any sequence will work. Here’s a fairly nice one that isn’t quite so familiar.

Let $X=\mathbb{R}\setminus\{2^{-n}:n\in\mathbb{Z}^+\}$ as a subspace of $\mathbb{R}$ with the usual topology, and let $Y$ be the quotient of $\mathbb{R}$ obtained by identifying $\mathbb{Z}^+$ to a point; the desired space is $X\times Y$. For $n,k\in\mathbb{Z}^+$ let $$F(n,k) = \{x\in X:|x-2^{-n}|\le 2^{-k}\}\times\{n-2^{-k}\},$$ and let $$F=\bigcup_{n,k\in\mathbb{Z}^+} F(n,k)\;.$$

The closure of $F$ in $\mathbb{R}^2$ is $F\cup\{\langle 2^{-n},n\rangle:n\in\mathbb{Z}^+\}$, so $F$ is closed in $X\times \mathbb{R}$, but $F$ is not closed as a subset of $X\times Y$: if $z\in Y$ is the image of $\mathbb{Z}^+$ under the quotient map, and $p = \langle 0,z\rangle$, it’s easy to see that $p$ is a limit point of $F$. However, $p$ is not the limit of any sequence in $F$.

To see this, suppose that $\sigma=\langle p_i:i\in\omega\rangle$ is a sequence in $F$, say with $p_i\in F(n_i,k_i)$ for $i\in\omega$, and suppose that $\sigma\to p$. If $\{n_i:i\in\omega\}$ were bounded, the projection of $\sigma$ on $X$ wouldn’t approach $0$, and if $\{k_i:i\in\omega\}$ were bounded, the projection of $\sigma$ on $Y$ wouldn’t approach $z$, so we might as well assume that $\langle n_i:i\in\omega\rangle$ and $\langle k_i:i\in\omega\rangle$ are strictly increasing. But then it’s easy to see that $\langle n_i-2^{-k_i}:i\in\omega\rangle \not\to z$ in $Y$, contradicting the assumption that $\sigma\to p$.

Thus, the space obtained from $X\times Y$ by isolating $p$ has exactly the same convergent sequences as $X\times Y$.

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Several nice examples have been given. I'll add a general construction, which can be used to produce more examples.

If $X$ is a space which is not sequential and $Y$ is the sequential coreflection of $X$, then $X$ and $Y$ have the same underlying set, they have the same convergent sequences and the topology of $Y$ is strictly finer than $X$.

The sequential coreflection of a topological space $X$ can be obtained in the following way:

1. Take all subspaces of $X$ of the form $\{x_n; n\in\mathbb N\}\cup\{x\}$ where $(x_n)$ is a sequence which converges to $x$ in $X$.
2. Make a direct sum of these spaces and then the quotient space obtained by identifying "all copies" of a given point $x$.

A picture would be better, but I hope with a little effort this construction can be understood from what I wrote. This is basically the construction given in Proposition 1.12 of S.P.Franklin: Spaces in which sequences suffice, Fund. Math. 57 (1965), 107-115.

The sequential coreflection can be equivalently described as the topology, in which closed sets are precisely the sequentially closed sets of the original topology. By sequentially closed we mean a set $A$ which for any convergent sequence of elements of $A$ contains all limits of this sequence.

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