Timeline for Connectedness of sequence spaces (countable products) in different metrics
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 2, 2019 at 13:47 | answer | added | Niels J. Diepeveen | timeline score: 1 | |
| May 27, 2019 at 12:25 | comment | added | user85365 | The space is in general not compact. For instance, if X = [0,1], consider the sequence $(\bar{x}^{(n)})_{n\in\mathbb{N}}$ in $X^{\mathbb{Z}}$, whose $n$th component is the sequence $x^{(n)}_i = (1 - \frac{1}{n})^i$. Clearly, for each fixed $i$, $x^{(n)}_i$ converges to $1$. Hence, if the sequence $(\bar{x}^{(n)})_{n\in\mathbb{N}}$ converges in $(X^{\mathbb{Z}},D_1)$, then it can only converge to the constant sequence all of whose components are $1$, since convergence in this space means uniform convergence. But the convergence is not uniform! Hence, the sequence has no limit points. | |
| May 26, 2019 at 11:31 | comment | added | Taras Banakh | But your space $X$ is compact and so is its countable power. | |
| May 24, 2019 at 14:11 | comment | added | user85365 | Thank you for this very helpful hint. The property that you define (call it uniform connectedness) seems to be a necessary condition for the connectedness of the sequence space with metric $D_1$, but maybe not sufficient. There is an exercise in the book "General Topology" by Engelking (Ex. 6.1.D) that indicates that $X^{\mathbb{Z}}$ should be compact to draw the desired conclusion that uniform connectedness of $X$ implies connectedness of $X^{\mathbb{Z}}$. | |
| May 23, 2019 at 13:36 | comment | added | Taras Banakh | I hope that the space $(X^{\mathbb Z},D_1)$ is connected if and only if for any $\varepsilon>0$ there exists $n\in\mathbb N$ such that any points $x,y$ in the metric compact space $(X,d)$ can be lined by a chain of points $x=x_0,\dots,x_n=y$ such that $d(x_{i},x_{i+1})<\varepsilon$ for every $i<n$. | |
| May 22, 2019 at 14:18 | comment | added | user85365 | No, that's actually not clear. In any case, the topology induced by the sup-metric is always finer than the topology induced by any of the metrics $D_2$: The identity map from $(X^{\mathbb{Z}},D_1)$ to $(X^{\mathbb{Z}},D_2)$ is continuous. There is an argument showing that it is strictly finer, but at the moment I don't remember it. | |
| May 22, 2019 at 12:57 | comment | added | YCor | Is it clear that the second topology does not depend on the choice of $D_0$? | |
| May 22, 2019 at 12:39 | history | edited | YCor | CC BY-SA 4.0 |
fixed formatting
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| May 22, 2019 at 12:28 | history | asked | user85365 | CC BY-SA 4.0 |