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Let $\tau_1$ and $\tau_2$ be two topologies on a non-empty set $X.$ Let the sequential convergence with respect to one of the topologies be equivalent to the sequential convergence with respect to the other. Can we conclude that $\tau_1 = \tau_2\ $?

I don't think that it is correct. Our instructor hinted that we can construct a counter-example in the space $\ell^1(\mathbb N),$ the space of all summable sequences of complex numbers. But I can't figure it out. Could anyone give me some hint?

Thanks a bunch!

RKC
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    In $\ell^1$ the weak and norm topology are distinct but have the same convergent sequences. This is known as the Schur property – Alessandro Codenotti Sep 02 '21 at 06:10
  • @Alessandro Codenotti$:$ How to show that in the space $\ell^1 (\mathbb N)$ weak convergence implies norm convergence? Also how to produce an open subset which is in the norm topology but not in the weak topology? – RKC Sep 02 '21 at 06:43

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Here's an easy example. For $\tau_1$ take the discrete topology on $\mathbb{R}$. For $\tau_2$ let every point be isolated, except $0$ whose neighbourhoods will be the co-countable sets. In both topologies the convergent sequences are those that are constant eventually.

hartkp
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There are simpler examples that do not require functional analysis at all.

The cocountable topology on an uncountable set has this property that a sequence is convergent if and only if it is eventually constant (see this). The same property applies to the discrete topology which is not only different (in the sense of literal equality) but also not homeomorphic to the cocountable one.

For $\ell^1$ space you are dealing with the Shur's property. To see how $\ell^1$ has it read this: $\ell^1$ Schur property

freakish
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  • I want a counter-example in $\ell^1(\mathbb N)$ @freakish. – RKC Sep 02 '21 at 07:24
  • What is $y_n^{(0)}$ in this answer$:$ https://math.stackexchange.com/a/42620/960468 – RKC Sep 02 '21 at 07:44
  • @RabinKumarChakraborty $\ell^1$ is a space of sequences. And in the answer $y^{(n)}$ refers to a sequence of sequences, i.e. $y^{(n)}\in\ell^1$ while $y^{(n)}_k$ is its $k$-th element. – freakish Sep 02 '21 at 12:54
  • Ok. I understand now. But the last part of the proof is still not clear to me. First of all I don't think the inequality at the fag end is correct. I have added a comment there in this regard. Also I have a doubt on the approximation which bounds the first part by $\frac {98} {100} \varepsilon.$ How does that $98$ come into the picture? – RKC Sep 02 '21 at 16:41