![]() ![]() The sequence is convergent so its Cauchy : there exists an N N such. Let x x be a limit of sequence (uk)k ( u k) k of points of F F. Kaplansky states the following on page 130 of Set theory and metric spaces: 'If the Tietze theorem admitted an easier proof in the metric case, it would have been worth inserting in our account. Assuming that the xi x i s are different one from each other, take the minimum of d(xi,xj) d ( x i, x j) for i j i j, noted. (I may, of course, simply be missing some simpler approach.)įor $t\in\Bbb R$ let $A_t=\\Big((na,nb)\cup(-nb,-na)\Big)$$ is a compact subset of $\Bbb R$. begingroup The question made me wonder whether there is a simpler proof of Tietzes extension theorem (generalizing Urysohn) for the case of metric spaces. Here’s an extended hint to get you started in a direction that will work, although the argument is a bit more complicated than I’d hoped. (You don’t need both $\sup A=+\infty$ and $\inf A=-\infty$.) However, this is more a matter of restating the problem than of actually having a direction in which to try to move. Showing that there this happens for at least one $t\in\Bbb R$ would indeed suffice, and you could do that if you could find such an $A$ for which either $\sup A=+\infty$ or $\inf A=-\infty$ (which I suspect is what you mean when you speak of them not existing). ![]() The negation would be, a metric space is not bounded if for every n N n N, there exist x. A metric space is bounded if there exists an M N M N such that (x, y) M ( x, y) M for all x, y X x, y X. Then show for all $x, y \in X$, $d(x, y) < R + 2$.It’s not entirely clear what you mean by ‘$D(A)$ does not exist’, but it appears that you mean that it’s infinite. My question is on the bolded part and if I am correct in making that statement. Let $R$ be the max of $d(x_i, x_j)$ as $i, j$ vary. Now the diameter (max distance between points) in $X$ does not just depend on the number of balls but also on the distances between the centers. Then you will end up with a finite open cover consisting of a finite number of balls of radius 1, $B(x_i, 1)$. We will now extend the concept of boundedness to sets in a metric space. But no one here understood it, because you didn't say what $\mathscr U$ is.Įdit: Again, regarding your original attempt, you should get rid of the variables $\epsilon_i$ by just taking all radii to be $1$. In Section 5 we prove Theorem B, that is, we define a product on Lip0(M) so that Lip0(M) becomes a Banach algebra when M is an unbounded metric space, and we. Therefore closed and bounded subset of a complete metric space need not be compact. Remark: If we assume in addition that the set is totally bounded, the usual proof can be carried out. Then E E is closed and bounded, but not compact. Assuming your $\mathscr U$ consists of one ball of some radius about each point of $X$, your proof is (with some editing) basically correct. This metric makes X X a complete metric space. ![]() You take it from here.īy the way, in your attempt, you violated the first rule of proof writing, which is every symbol must be explained at the point where it is introduced. ![]() A set in is bounded iff it is contained inside some ball of finite radius (Adams 1994). Let $X$ be an unbounded metric space and assume towards a contradiction that $X$ is compact. A set in a metric space is bounded if it has a finite generalized diameter, i.e., there is an such that for all. I feel like the end of my proof is obvious, but I cant explain it. Is an infinite set with no limit point unbounded in an arbitrary metric space Hot Network Questions Can I do assembly programming using the kit I bought or do I have to get another setup Expanding CamelCase for readability using fontspec information Connecting double balanced TRS outputs to TRS unbalanced input. I am having a difficult time explaining the result. ![]()
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