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.

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    $\begingroup$ 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. $\endgroup$
    – Squirtle
    Sep 2 '15 at 14:48
  • $\begingroup$ $\bar{\mathbb{R}}$ 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 $\infty$. $\endgroup$
    – Falko
    Sep 2 '15 at 14:53
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    $\begingroup$ @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 $p \neq q; d(p,p) = 0$ ..." $\endgroup$
    – Falko
    Sep 2 '15 at 15:02
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    $\begingroup$ 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. $\endgroup$
    – Squirtle
    Sep 2 '15 at 15:06
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    $\begingroup$ @Squirtle Using $\arctan$, the extended real line is homeomorphic to $[-\pi/2, \pi/2]$. You can use that to put a metric on it. $\endgroup$ Sep 2 '15 at 15:19

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.

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    $\begingroup$ 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 $[0,1]$). $\endgroup$ Sep 2 '15 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – Aloizio Macedo
    Sep 2 '15 at 15:25
  • $\begingroup$ 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. $\endgroup$ Sep 2 '15 at 19:43
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    $\begingroup$ 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). $\endgroup$
    – Procore
    Oct 13 '16 at 4:02
  • $\begingroup$ 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. $\endgroup$
    – Procore
    Oct 14 '16 at 1:14

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