![]() Then show for all $x, y \in X$, $d(x, y) < R + 2$. 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, $B(x_i, 1)$. Define a metric d on X by d(x,y) 0 if x y, and d(x,y) 1 if x neq y. 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 $1$. Assuming your $\mathscr U$ consists of one ball of some radius about each point of $X$, your proof is (with some editing) basically correct. ![]() If f ( x) B for all x in X, then the function is said to be bounded (from) below by B. The setting considered in this paper is a complete metric. citation needed If f is real-valued and f ( x) A for all x in X, then the function is said to be bounded (from) above by A. This latter definition is suitable for generalization to metric spaces, using p-weak upper gradients. 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. 1 A function that is not bounded is said to be unbounded. (Thus, X is a complete metric space with the metric where ( x, x ) x - x. Let $X$ be an unbounded metric space and assume towards a contradiction that $X$ is compact. I feel like the end of my proof is obvious, but I cant explain it. ![]() ![]() I am having a difficult time explaining the result. ![]()
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