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In Definition $2.18$ of Baby Rudin, Rudin defines boundedness for metric spaces as follows: given a metric space $X$, and a set $E \subset X$, we say $E$ is bounded if there exists a real number $M$ and a point $q \in X$ such that $d(p,q) < M$ for all $p \in E$.

Per this definition, if $E=X=\varnothing$, then $E$ is unbounded. The reason is that "there exists a point $q \in X$" is a false statement if $X = \varnothing$.

Should this be regarded as a minor mistake/typo in Baby Rudin? Or are there particular reasons the empty set should be considered unbounded?

If we use the alternative definition that a set $E$ is bounded if there exists a real number $M$ such that for all $p,q \in E$, $d(p,q) < M$, then this coincides with the above definition for nonempty sets, but also always considers the empty set bounded.

MathematicsStudent1122
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    Does this answer your question? Why is the empty set bounded? – Rushabh Mehta Feb 16 '20 at 03:08
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    @DonThousand No, it doesn't answer the question, because I'm specifically inquiring about Rudin's definition of bounded, which is such that the empty set is unbounded per that definition. I am aware of the numerous arguments in favor of why the empty set should be considered bounded. – MathematicsStudent1122 Feb 16 '20 at 03:10
  • whether or not you consider the empty set bounded is more a matter of pedagogy and ease in stating things. I wouldn't really call it a "flaw" with Rudin; I would call it a nonstandard convention. That is, unless Rudin claims at some point that something holds only for bounded sets and it in fact holds for the empty set as well – Brevan Ellefsen Feb 16 '20 at 03:11
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    Your concern seems to be that the definition of "bounded" breaks down when not only the set $E$, but the entire metric space $X$ is the empty set. (We can still perfectly well say that the empty set is bounded in the metric space $\mathbb{R}$.) I'm not sure there's much that needs to be said about a metric space with no points at all. – Gregory J. Puleo Feb 16 '20 at 03:11
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    @DonThousand: Actually, if you look closely, this is not a duplicate. And, indeed, Rudin's definition is not good. One way to modify it is as you say, require the diameter of $M$ to be finite. Another is to say for *for every $q\in X$ there exists $M$ such that for all $p\in M$, $d(p,q)<M$. – Moishe Kohan Feb 16 '20 at 03:12
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    @GregoryJ.Puleo That should be an answer. – Noah Schweber Feb 16 '20 at 03:13
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    @BrevanEllefsen It's not a major issue, but it makes some theorems strange. For instance, the theorem "if $X$ is a compact metric space, then $X$ is bounded" breaks down if $X$ is empty. – MathematicsStudent1122 Feb 16 '20 at 03:13

1 Answers1

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This definition is not standard and should be considered as an error in Baby Rudin. There is no reason whatsoever to consider the empty set unbounded in the empty space, and Rudin almost certainly did not intend to do so. Indeed, I would not be surprised if there are some theorems later in the text that are incorrect with his definition.

Eric Wofsey
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  • Rudin's definition of boundedness for a subset of a metric space seems to agree with Wikipedia, https://en.wikipedia.org/wiki/Bounded_set – Coriolanus Feb 16 '20 at 03:56
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    So, Wikipedia has the same mistake. It's an easy mistake to make. Quite possibly the definition on Wikipedia was even taken from Baby Rudin (there is no source listed and neither of the references at the end of the article appear to be a source for it). – Eric Wofsey Feb 16 '20 at 04:01