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:
- Is there a more topological/simpler example?
- 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
Best Answer
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.