Do the same sequential convergence in two topologies the result are the same topologies? Let $X$ be topological space with topologies $\tau$, $\tau'$. Let $(x_{n})\in X$ and $x\in X$ and $(x_{n})\rightarrow x$ with $\tau$ if and only if $(x_{n})\rightarrow x$ with $\tau'$. Do the two topologies on $X$ coincide?
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What conditions must be x to the question above is true? – user244115 Jul 20 '15 at 12:07
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This is my first instinct upon reading this problem: Assume that there is a $y\in X$ and $U\in\tau$ such that $y\in U$ and there is no $V\in \tau'$ such that $y\in V\subseteq U$ (this is what it means that the two topologies do not coincide, although technically it might be with $\tau$ and $\tau'$ reversed). Now make a sequence that converges (to $y$) in one topology but not in the other. – Arthur Jul 20 '15 at 14:53
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1What have you attempted? What in particular do you need assistance with? – anak Jul 20 '15 at 14:53
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We know the same sequential convergence in two topologies dont result topologies are the same. We know the same sequential convergence in two topologies dont result topologies are the same. Nonetheless, in the case that $(X,\tau)$ and $(X,\tau')$ are both first-countable (for example metrizable) , necessarily $\tau \tau'$ . But I cant prove it. – user244115 Jul 20 '15 at 15:05
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Perhaps give a metric to reach of the topologies and show that an open (or closed) ball in one is still open (or closed) in the other. – Arthur Jul 20 '15 at 15:08
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What @bburGsamohT says. It is against the rules to ask the same question many times. Keep editing the first question. And wait longer. – Jyrki Lahtonen Jul 20 '15 at 15:59
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See also: Example of different topologies with same convergent sequences – Martin Sleziak Jul 21 '15 at 07:55
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About additional question in your comment: You basically need to know that every first-countable space is sequential. – Martin Sleziak Jul 21 '15 at 08:02
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Fix any closed subset $U \subseteq X$ of the topology $\tau$. I will show that $U$ must also be a closed set of $\tau'$.
Well, suppose $x$ is a limit point of $U$ in the topology $\tau'$. Then, there is a sequence $(x_n) \subseteq U$ such that $x_n \to_{\tau'} x$ (in $\tau'$). But, by the hypothesis in your question, $x_n \to_\tau x$ also (in $\tau$). This implies that $x \in U$ because $U$ is closed in $\tau$. So, $U$ contains each of its $\tau'$-limit points i.e. it is closed in $\tau'$.
By exactly the same argument with $\tau$ and $\tau'$ interchanged, we also get that if $U$ is closed in $\tau'$, it is also closed in $\tau$. Thus, the two topologies must be the same.
EDIT:
My answer above was meant for a related but different question by the same user. For technical reasons, the moderators deleted that question and they decided to "paste" my answer onto this older site with a related question by the same user. As such, this answer is no longer correct.
The question I answered had the added hypothesis that $\tau$ and $\tau'$ are metrizable. This assumption is necessary to characterize limit points (and hence, closed sets) in terms of sequences which is a crucial step in my arguments.
For actual answers to this particular question, look at the older answers below.
balddraz
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Not necessarily. Let $X=\omega_1+1$, let $\tau$ be the order topology on $X$, and let $\tau'$ be the finer topology obtained by isolating the point $\omega_1$. Then $\langle X,\tau\rangle$ and $\langle X,\tau'\rangle$ have the same convergent sequences, but they are not homeomorphic: the first is compact, and the second isn’t.
If you’re not familiar with transfinite ordinals, a simpler version of the same idea is to let $X$ be any uncountable set. Fix a point $p\in X$, and let
$$\tau=\{U\subseteq X:p\notin U\text{ of }X\setminus U\text{ is countable}\}\;.$$
Let $\tau'$ be the discrete topology on $X$. Clearly $\langle X,\tau\rangle$ and $\langle X,\tau'\rangle$ are not homeomorphic, since $p$ is not isolated in $\langle X,\tau\rangle$, but they have the same convergent sequences: in each of the only convergent sequences are the sequences that are eventually constant.
Added: If both topologies are sequential, then they are the same, since in that case the topology is completely determined by the convergent sequences. In particular, all first countable (and hence all metrizable) spaces are sequential.
Brian M. Scott
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@user244115: If both topologies are metrizable, then they are the same topology: in a metric space the convergent sequences determine the topology. (In fact you need much less than metrizability: the topologies need only be sequential. A Banach algebra is metrizable. – Brian M. Scott Jul 20 '15 at 12:14
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No. Consider $X=\Bbb{R}$, $\tau$ the indescrete topology, $\tau'$ the cofinite topology. Then Every sequence converges to every point in both topologies, but these two are distinct.
Nonetheless, in the case that $(X, \tau)$ and $(X, \tau')$ are both first-countable (for example metrizable) , necessarily $\tau = \tau'$. To show this, one can easily use the argument caracterizing closed sets as the sets closed under limit points of sequences.
Crostul
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This question does not make any sense, since we are talking about two topologies. – Crostul Jul 20 '15 at 12:38
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In $\ell^{1}$ if $(x_{n})\rightarrow x$ in the weak topology then $(x_{n})\rightarrow x$ in normed topology, but not coincide two topologies. – user244115 Jul 20 '15 at 13:03
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Thank you. Sorry for not being able to prove that for case they are both first-countable .Please can you help me? – user244115 Jul 20 '15 at 13:43
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The indiscrete topology is the unique topology for which every sequence converges to every point. If $\tau$ is a topology on $X$ which is not indiscrete, find $U\in \tau$ which is nonempty and not $X$, find $x\in U$ and $y\in X\setminus U$, then the constant sequence $w_n=y$ never enters $U$ and therefore cannot converge to $x$. – Jade Vanadium Jul 28 '22 at 19:59