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Note that we are working in the reals, not the extended reals.

Do you understand a closed interval as "an interval that is a closed set" or as "an interval that includes both its endpoints"? If the former, then you would consider $(-\infty,\infty)$ to be a closed interval. If the latter, then you would not consider $(-\infty,\infty)$ to be a closed interval.

I belong to the latter camp, but I would like to get some consensus here, if possible. (I have been graded down for this, have asked around, and I am surprised that everyone I've consulted calls $(-\infty,\infty)$ a closed interval. Am I really crazy?)

p.s. To be very clear, I am not asking if $(-\infty,\infty)$ is a closed set -- it sure is.

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    $\begingroup$ There is another (better) definition saying that an interval (or more genreally any subset) is closed if it contains all its limit point. So in this case $(-\infty, \infty)$ is also closed. $\endgroup$ – user99914 Oct 29 '13 at 5:39
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    $\begingroup$ @John Your definition is equivalent to the first one I gave (since closed sets are precisely the sets that include all their limit points). $\endgroup$ – Ryan Oct 29 '13 at 5:45
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    $\begingroup$ @Ryan: If an interval is a closed set then it is a close-interval! because an interval is either open-interval, hence an open set , otherwise a one-sided open-interval which is not a closed-set. $\endgroup$ – Souvik Dey Oct 29 '13 at 6:00
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    $\begingroup$ You have already answered your own question. If one accepts the first definition, then $\Bbb R$ is a closed interval. If one accepts the second definition, then $\Bbb R$ is not a "closed interval," although it is an interval that is a closed set. Both definitions are offered in practice. However the second definition creates the unfortunate and unintended circumstance that a closed interval is not a "closed interval." Thus, it is more technically sensible to accept the first definition. In practice, I believe the first is used more often by the more careful authors. $\endgroup$ – anon Oct 29 '13 at 6:01
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    $\begingroup$ @Ryan It seems to me that the all the comments and answers understand closed interval as equivalent to closed set. If you're looking for consensus, or for which definition is more broadly used then I think you have your answer. $\endgroup$ – in_mathematica_we_trust Oct 29 '13 at 6:10
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Amazing degree of consensus on the wrong answer. I won't bother giving my opinion (which I sort of hinted at in the previous sentence) because I am nobody and my opinion means nothing. Instead I will suggest a method you can use to get an answer, which may be slightly more valid than polling a handful of random internet addicts.

Go to the library and find the shelf with all the introductory analysis books. Look up the statements of theorems like "a continuous function on a finite closed interval attains its maximum" and "a continuous function on a finite closed interval is uniformly continuous". Count how many books include the word "finite" and how many omit it. I haven't done this myself, so I have no idea what result you will get.

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  • $\begingroup$ I think the number of books that will omit the word "finite" when discussing theorems that need compact sets should be the empty set, thus the complement, this discussion, is closed. jk $\endgroup$ – Brady Trainor Oct 29 '13 at 7:00
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    $\begingroup$ @BradyTrainor Funny, somebody suggests an empirical approach to solve a question and you reply instantly that you do not need to do the experience because you know the answer. What happens if the data contradicts you? The data is wrong? $\endgroup$ – Did Oct 29 '13 at 8:39
  • $\begingroup$ Well, when I hear "closed interval" I think [a,b]. But what do I know? $\endgroup$ – bof Oct 29 '13 at 10:24
  • $\begingroup$ Precisely. I could accept allowing (−∞,∞) to be called a closed interval (since it is apparently not useful to insist otherwise, esp. in topology), but I would also point out that the notation indicates otherwise. Your suggestion to consider the extreme values theorem on the reals is excellent and convincing; you're the only person I've asked this question of today (not just on this website) to agree with the answer that I originally had thought was clear as ink. Apparently the earth is flatter than I had assumed. $\endgroup$ – Ryan Oct 29 '13 at 10:35
  • $\begingroup$ BTW by this reasoning $[0,-1]$ is not a closed interval. $\endgroup$ – sdcvvc Oct 29 '13 at 11:06

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