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

Solution 1:

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.

Solution 2:

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:

Solution 3:

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 $X\times Y$.