Talk:Extended real number line

message: I question the notation in this article.  should the symbol for the extended reals be R?  that's the same symbol for the regular real line.  using the same symbol for both could be endlessly confusing.  None of my textbooks (Rudin, Royden, Schechter) uses this notation, but they also don't seem to have a prefered notation.  They just call it by name.  I think this article should follow that same convention.  -Lethe | Talk
-- It isn't R, but R with a bar on top.
::Which displays for some people, including me, as just R.  Dbenbenn 00:02, 25 Dec 2004 (UTC)
thread_title:  notation 

message: It is currently stated that "a × (b + c) and (a × b) + (a × c) are either equal or both undefined." but unfortunately this is incorrect. Consider a = +oo, b = 2 and c = -1, for example. Then a × (b + c) is defined, but (a × b) + (a × c) is not.
I'm not sure how the entry should best be corrected.
:Good catch!  I think the following modification is true
:*a × (b + c) and  (a × b) + (a × c) are equal if both are defined.
:Dbenbenn 02:15, 29 Dec 2004 (UTC)
thread_title:  Failure of distributive property 

message: :f_n(x) = \begin{cases} 2n(1-nx), & \mbox{if } 0 \le x \le \frac{1}{n} \\ 0, & \mbox{if } \frac{1}{n} < x \le 1\end{cases}
This is not the example I want. This is an example to show that you can't interchange the limits at will. I want an example to show that it's OKAY to get the limit function to take on ∞ at one point say, and still satisfy one of the convergence theorems. I can't come up with the right example at the moment, maybe my brain is fried. If anyone remembers an example like this, feel free to put it there. It's one of those standard examples, I just can't think of it. Revolver 19:31, 16 September 2005 (UTC)
thread_title: not the example I want

message: This article seems to play fast and loose with the definition of "continuous".  Under the normal definition of continuity, small changes in x should produce relatively small changes in y.  This is clearly not the case where x=0, where changes in y become unbounded.  The function is continuous by the topological definition, but this is not clear cojoco 03:47, 6 November 2007 (UTC)
replies: text: Which other definition is there? FilipeS (talk) 02:38, 30 November 2007 (UTC)
replies: text:  There are at least two definitions: under the first definition in Continuous_function, this function is not continuous, because nearby differences become unbounded at x=0.  Under topological definition in the same article, it is.  The statement "the function y=x^{-2} can be made continuous by setting the value to +∞ for x = 0" is misleading because this is not true under the normal definition of continuity, so the statement should be clarified.  So, I've clarified it.cojoco (talk) 21:21, 21 December 2007 (UTC)

text: You are right, in so far as in the standard definition of continuity, when we say that "the limit of f(x) as x approaches c must exist and be equal to f(c)", we assume that f is a real-valued function. Thus, f(c) cannot be infinite, by definition. FilipeS (talk) 17:12, 11 January 2008 (UTC)
replies: text: But in the extended reals, the limit does apporach +∞, which is the value added.--Hagman-de (talk) 20:25, 27 July 2010 (UTC)
thread_title:  "the function <math>y=x^{-2}</math> can be made continuous by setting the value to +∞ for x = 0" 

message: I'm a little confused by the formal Cauchy definition of +∞. If you take the sequences \{ 1, 2, 3, \ldots \} and \{ 2, 3, 4, \ldots \}, then these represent +∞, but their differences is not a null sequence.  —Preceding unsigned comment added by Jweimar (talk • contribs) 01:12, 26 February 2009 (UTC)
replies: text: \bar{\R} is not a metric space (at least not by simply extending the standard metric). One can define a metric on it, but that is not translation invariant, so ultimately the difference of two convergent sequences need not be a null sequence.--Hagman-de (talk) 20:22, 27 July 2010 (UTC)
thread_title:  Cauchy definition 

message: This example doesn't seem right to me: \sin x^{-1} doesn't have a limit (at x=0, which I assume is meant) in the first place, so it's not interesting that its reciprocal doesn't either.  Perhaps we want the function x\sin x^{-1}, which vanishes at 0 but whose reciprocal fails to have a limit in \overline\mathbb R because it changes sign infinitely often?  --Tardis (talk) 18:29, 1 July 2009 (UTC)
thread_title:  The 1/sin(1/x) example 

message: "The expression 1/0 is not defined either as +∞ or −∞..." Is this correct? The wolfram reference says |x/0| = +∞ when x is not zero. Is there an important role the absolute value plays there? (http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html)  —Preceding unsigned comment added by KarateClown (talk • contribs) 01:33, 27 November 2010 (UTC)
thread_title:  Definition of 1/0 

message: Can anyone extend the "Arithmetic operations" section of this article to include exponentiation?? Georgia guy (talk) 13:47, 23 October 2014 (UTC)
thread_title:  Exponentiation 

message: The article says:
The expression 1/0 is not defined either as +∞ or −∞, because although it is true that whenever f(x) → 0 for a continuous function f(x) it must be the case that 1/f(x) is eventually contained in every neighborhood of the set {−∞, +∞}, it is not true that 1/f(x) must tend to one of these points. An example is f(x) = (sin x)/x (as x goes to infinity). (The modulus | 1/f(x) |, nevertheless, does approach +∞.)
The example given implies that sin(infinity) is 1. However, the truth is that sin(infinity) cannot be defined. Any flaws in the thinking I'm using?? Georgia guy (talk) 21:26, 7 October 2015 (UTC)
replies: text: It probably ought to say "as x goes to zero" instead. Double sharp (talk) 04:29, 27 October 2015 (UTC)
replies: text: No, wait, it's correct; I just misread it. The whole point of this example is that sin x changes sign infinitely often as x approaches infinity, so that even though (sin x)/x approaches 0 as x gets large, its reciprocal does not approach either positive or negative infinity. Double sharp (talk) 04:39, 12 November 2018 (UTC)
thread_title:  Anyone able to test faulty info in this?? 

message: Per this book, "each subset of \overline{\mathbb{R}} has both supremum and infimum". Therefore, this revert was inappropriate. Ladislav Mecir (talk) 16:34, 8 December 2019 (UTC)
replies: text: The empty set is a subset of \overline{\mathbb{R}} and it has no supremum or infimum. Some sources are a bit sloppy. Let's base our articles on less sloppy ones . - DVdm (talk) 16:40, 8 December 2019 (UTC)
replies: text: In this case, the sources are what we should respect. Can you cite any source that contradicts this claim? Ladislav Mecir (talk) 16:43, 8 December 2019 (UTC)

text: BTW, the sources are pretty specific that the supremum of the empty set in \overline{\mathbb{R}} is  and the infimum of the empty set is . Ladislav Mecir (talk) 16:47, 8 December 2019 (UTC)
replies: text: I did a book search and found as many sloppy ones as less sloppy ones. Let's just go with the latter... - DVdm (talk) 16:55, 8 December 2019 (UTC)
replies: text: Do I understand correctly that you do not care to mention any source, yet you insist that there are some "not sloppy ones" that contradict the claim that "each subset of \overline{\mathbb{R}} has both supremum and infimum"? Ladislav Mecir (talk) 17:04, 8 December 2019 (UTC)
replies: text: It is true that the property is true for the empty set, but removing the restriction without explanation may confuse many readers. So, I have self reverted my revert, and added such an explanation. D.Lazard (talk) 17:44, 8 December 2019 (UTC)
replies: text: Thank you. In my opinion, your addition explaining what the infimum and supremum of the empty set are is useful. Ladislav Mecir (talk) 18:00, 8 December 2019 (UTC)
replies: text:  I concede. Any number is at the same time smaller and larger than any number in the empty set. - DVdm (talk) 18:25, 8 December 2019 (UTC)
thread_title:  Every subset of the extended real line has both a supremum and an infimum