For all \( n \in \mathbb{N} \) and for all \( x_i, \; i=1,2,\dots, n \) there exists \( z \in X \) such that \( \displaystyle \frac{1}{n} \sum_{i=1}^{n} d(z, x_i) = r \).
Rendezvous value
- Tolaso J Kos
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Rendezvous value
Let \( (X, d) \) be a complete and a connected metric space. Prove that there exists a unique number \( r=r(X, d)>0 \) with the property:
For all \( n \in \mathbb{N} \) and for all \( x_i, \; i=1,2,\dots, n \) there exists \( z \in X \) such that \( \displaystyle \frac{1}{n} \sum_{i=1}^{n} d(z, x_i) = r \).
For all \( n \in \mathbb{N} \) and for all \( x_i, \; i=1,2,\dots, n \) there exists \( z \in X \) such that \( \displaystyle \frac{1}{n} \sum_{i=1}^{n} d(z, x_i) = r \).
Hidden Message
Imagination is much more important than knowledge.
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Re: Rendezvous value
Hi Tolis,
I believe we need some boundedness condition. For example the result is not true for $\mathbb{R}$.
I believe we need some boundedness condition. For example the result is not true for $\mathbb{R}$.
- Tolaso J Kos
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Re: Rendezvous value
Hi Demetres,
how about if we replace complete with compact? I think that the exercise now would be correct. I was thinking about this in the last $3$ days and I came up to the same conclusion that this cannot hold in $\mathbb{R}$ as you said. But I think I have a huntch that replacing complete with compact may actually work. What do you think?
how about if we replace complete with compact? I think that the exercise now would be correct. I was thinking about this in the last $3$ days and I came up to the same conclusion that this cannot hold in $\mathbb{R}$ as you said. But I think I have a huntch that replacing complete with compact may actually work. What do you think?
Imagination is much more important than knowledge.
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Re: Rendezvous value
Yes compactness includes a boundedness condition so it now looks more plausible. I don't have a proof yet, neither for the existence nor for the uniqueness. I'll need to think about it.
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