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I've got a question reading the demonstration of the Theorem 3.2 in POMA of Rudin. Indeed, he says that every convergent sequence in a metric space is bounded. My question is:

Is $\bar{\mathbb{R}}$ with the usual distance a metric space?

Indeed, the sequence $(u_n)_{n \in \mathbb{N}}$ defined by $u_0 = +\infty$ and then $u_n = \frac{1}{n} \forall n \geq 1$ is convergent but not bounded... Thus I guess that it is not.

Najib Idrissi
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yarmenti
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    Extended real numbers are weird.... Because we allow $\infty$ to be in this number system it doesn't seem too strange that every sequence is bounded since after all $-\infty \le r \le \infty$ for every $r\in\overline{\mathbb{R}}$. So I would say that this sequence is bounded in this regard.... but ask your professor since this is kind of a technical question. – Squirtle Sep 02 '15 at 14:48
  • $\bar{\mathbb{R}}$ with the usual distance is not a metric space, since a metric should only take nonnegative real numbers as values. But the distance between $\infty$ and some real number would be $\infty$. – Falko Sep 02 '15 at 14:53
  • for @Elzee: where do you find the mention that "only nonnegative real numbers" can define distance? – yarmenti Sep 02 '15 at 14:56
  • for @Squirtle: I understand your point, saying that the $\infty$ distance between 2 points belongs to $\overline{\mathbb{R}}$ and thus, it's ok. – yarmenti Sep 02 '15 at 14:57
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    @yarmenti: It should say so in the definition of a metric space. Let me cite from POMA (third edition, definition 2.15): "...is said to be a metric space if with any two points $p$ and $q$ of $X$ there is associated a real number $d(p,q)$, called the distance from $p$ to $q$, such that (a) $d(p,q) > 0$ if $p \neq q; d(p,p) = 0$ ..." – Falko Sep 02 '15 at 15:02
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    Also, it is possible to make the extended reals a metric space. You have to define a metric using an integral and the log function... but I can't remember the details off of the top of my head. In particular, the distance between $-\infty$ and $\infty$ is actually finite, so it defines a proper metric. – Squirtle Sep 02 '15 at 15:06
  • @Elzee. Ok, understood from his def. Thank you both of you. – yarmenti Sep 02 '15 at 15:13
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    @Squirtle Using $\arctan$, the extended real line is homeomorphic to $[-\pi/2, \pi/2]$. You can use that to put a metric on it. – Najib Idrissi Sep 02 '15 at 15:19
  • Ah yes... that's it! – Squirtle Sep 02 '15 at 15:22

1 Answers1

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No, $\overline{\mathbb{R}}$ with the usual distance is not a metric space. It is better understood as a topological space (with the order topology).

Although it is not a metric space with the usual distance, it is metrizable. Think about a way to put a metric on it.

But I will disagree with one of the comments (the one that says that $\overline{\mathbb{R}}$ is weird) and will present some arguments for that:

  • $\overline{\mathbb{R}}$ is compact. This is a great property. This also provides some more insight into the Bolzano-Weierstrass: EVERY sequence in $\overline{\mathbb{R}}$ has a convergent subsequence. This says that every sequence has either a subsequence converging to a real number, or a subsequence converging to some $\pm \infty$.
  • Every non-empty subset of $\overline{\mathbb{R}}$ has a $\sup$.
  • $\overline{\mathbb{R}}$ clears up the "converging to $\infty$" context, which is generally badly explained by a first course in real analysis. The definition for "converges for $x \rightarrow \infty$" or "converges to $\infty$" is seemed as artificial, and some people even say that "this is not a true convergence", and will even state: We will say $x_n \rightarrow +\infty$, but the sequence actually does not converge.
  • $\limsup$ does not have trivialities in his definition. For example, it is common to define that $\limsup$ is the $\sup$ of the set of numbers which are limits of subsequences of the given sequence. By the first item, this set is always non-empty. And by the second item, we always have the $\sup$. This happens naturally, and one doesn't need to define things artificially.
  • It even helps to prove that every continuous bijective function on an interval is an homeomorphism.

In my opinion, $\overline{\mathbb{R}}$ is not only not weird, but actually the right place to study analysis.

I made a blog post some time ago about this very subject.

Aloizio Macedo
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    The topological space $\bar{\mathbb{R}}$ is simply homeomorphic to the segment $[0,1]$. Not weird at all (and I don't think it needs such a lengthy explanation, if these facts are known for $[0,1]$). – Najib Idrissi Sep 02 '15 at 15:17
  • I aggree that being homeomorphic to $[0,1]$ is an argument for it being not-weird, but I don't agree with the last part of your sentence. For instance, the cantor set as a subset of $\mathbb{R}$ is homeomorphic to ${0,1}^{\mathbb{N}}$ (${0,1}$ with the discrete topology), but looking at the cantor set as a subset of $\mathbb{R}$ is often useful, Not only that, but the homeomorphism with $[0,1]$ does not make transparent (at least to me) how to prove that every continuous bijective function on an interval is an homeorphism, whereas $\overline{\mathbb{R}}$ does. – Aloizio Macedo Sep 02 '15 at 15:25
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    Your third point is somewhat tendentious: in an elementary course in which one is working strictly in $\Bbb R$ it is perfectly correct to say that a sequence like $\langle n:n\in\Bbb N\rangle$ diverges and to introduce the notation $\to\infty$ to describe a specific kind of divergence that is usefully distinguished from other kinds. – Brian M. Scott Sep 02 '15 at 19:43
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    Great discussion! Your saying $\mathbb{R}^{}=\mathbb{R}\cup{-\infty,+\infty}=[-\infty,+\infty]$, or the extended real numbers, is compact, complete, that the $\limsup$ (resp., $\liminf$) is well-defined in this set, is homeomorphic to an interval, and clears up ambiguity with $\pm\infty$. A quick proof to your answer is the pair $(\mathbb{R}^{},d)$, where $d:\mathbb{R}^{}\times\mathbb{R}^{}\rightarrow\mathbb{R}$ is the standard metric on $\mathbb{R}$, is $not$ a metric space as $+\infty-(+\infty)$ is not defined, (namely, not real...where it must be if we assume to the contrary). – Procore Oct 13 '16 at 04:02
  • I'll also add, in lieu of Elzee's comment w.r.t. the OP's question, for any $r,s\in\mathbb{R}^{}$, we have $d(r,s)$ is $not$ necessarily a real number (and need $not$ lie in $\mathbb{R}^{}$ as well), so $(\mathbb{R}^{},d)$ can't be a metric space, where $d$ is the standard metric on $\mathbb{R}$, and, for the sake of brevity, $+\infty−(+\infty)$ is not an extended real number. Lastly, define the metric $d_{0}$ on $\mathbb{R}^{}$ such that for $u,v\in\mathbb{R}^{}$ we have $d_{0}(u,v)=\big|\tan^{-1}(u)-\tan^{-1}(v)\big|$ giving $(\mathbb{R}^{}!!,d_{0})$ is a, complete, metric space. – Procore Oct 14 '16 at 01:14
  • The importance of \bar\textbb{R} is that it is compact, so many nice properties on finite sets can be transferred to it. Terrance Tao has an article about compactification on it. – user5280911 Jun 07 '20 at 16:12
  • I disagree with the idea of $\overline{\mathbb{R}}$ being the right place to study analysis. First of all, it doesn't have the structure of a field which is absolutely necessary for analysis. Second of all, analysis is usually taught during the first year. The potential infinity $\infty$ already causes a lot of confusion to beginners and encouraging them to treat it on the same terms as a usual number (which saying "let's study analysis in $\overline{\mathbb{R}}$ does) would make it even worse. – Adayah Jun 29 '22 at 17:27
  • That's why I think it is much better to first distinguish the neat behaviour of bounded closed intervals (Bolzano-Weierstrass, order completeness, compactness etc.) from the less regular unbounded intervals. Only then can we show the order-preserving homeomorphism between $\overline{\mathbb{R}}$ and $[0, 1]$, which makes it absolutely clear that topological and order properties are preserved while metric and algebraic properties are not. – Adayah Jun 29 '22 at 17:34
  • @AloizioMacedo your book for analysis is amazing . – Meet Patel Jul 24 '23 at 15:24
  • @AloizioMacedo Can you give me hint to prove "It even helps to prove that every continuous bijective function on an interval is an homeomorphism." – Meet Patel Dec 20 '23 at 12:32