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I am asking this because I have to do a counter example for "if the function is continuous on an interval then it's bounded on the interval." to that I can take $f(x) =x$ and show that $f$ is continuous on $\mathbb{R}$, but not bounded. For that I need to know that $(-\infty, \infty)$ is an interval.

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  • $\begingroup$ This is pretty vague. If a function is continuous on a closed interval, then it's bounded. That's a theorem. Any counterexample would have to use an open or semiclosed interval, then. I suppose you could count $(-\infty,\infty)$ as an interval, but there are counterexamples that use a bounded open interval, if you're trying to avoid $\infty$. $\endgroup$ – Benjamin Roycraft Dec 11 '14 at 9:40
  • $\begingroup$ Yes, it is an interval (in most definitions), but it is a good idea to find an example on a bounded interval. Try to think of a function that blows up near zero (doesn't have to be defined at zero). Then take that function on the interval $(0,1)$ or something of the kind. $\endgroup$ – Joonas Ilmavirta Dec 11 '14 at 9:42
  • $\begingroup$ It's a section of the real number line which contains every intervals you could come up with. I think instead of asking if $(\infty,\infty)$ is an interval, you should ask if $(\infty,\infty)$ is usually defined to be an interval. Since it's not obvious to you or me, it is a matter of conventional terminology. More so than, say, to ask if 7 is a natural number - where we break all of the other core conceptions if we negate it. $\endgroup$ – Nikolaj-K Dec 11 '14 at 9:43
  • $\begingroup$ Thank you friends. Now i got the idea. Joonas llmavirta thank you. $\endgroup$ – brjathu Dec 11 '14 at 9:51
  • $\begingroup$ @BenjaminRoycraft: "If a function is continuous on a closed interval, then it's bounded." That statement assumes that your definition of "interval" is "bounded interval." Because if you consider the closed interval $[0, \infty)$, say, then that statement is false. $\endgroup$ – Jesse Madnick Jun 21 '15 at 0:02
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I think it is worth mentioning the definition of an interval so you can see that $(-\infty, \infty)$ satisfies the criteria.

A subset $I$ of $\mathbb{R}$ is an interval if whenever $x, y \in I$, and $x < z < y$, then $z \in I$.

If $x, y \in (-\infty, \infty)$ and $x < z < y$, is $z \in (-\infty, \infty)$? Of course it does because $(-\infty, \infty)$ contains every real number, in particular $z$. Therefore, $(-\infty, \infty)$ is an interval.

As Martin Brandenburg mentions in a comment on his answer, one can characterise the intervals in $\mathbb{R}$ as the connected subsets. Note however that not every interval is of the form $(a, b)$; see this list.


More generally, one can define intervals in exactly the same way for any totally ordered set.

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Yes, $(-\infty,\infty)$ is an interval. But you will also find examples on the interval $(0,1]$.

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  • $\begingroup$ Considering the nature of the question, I would think it more probable that 'interval' refers to a finite open interval. $\endgroup$ – Jonas Dahlbæk Dec 11 '14 at 9:58
  • $\begingroup$ $(0,1)$ and $(-\infty,\infty)$ are homeomorphic, it doesn't really matter at all. Intervals can be defined as the connected subsets of $\mathbb{R}$, and $\mathbb{R}$ is connected. $\endgroup$ – Martin Brandenburg Dec 11 '14 at 10:15
  • $\begingroup$ It may not matter to you or me, but it certainly matters to a student learning the material for the first time! Anyway, I somehow failed to notice that you suggested to find a counterexample on a bounded interval in any case. $\endgroup$ – Jonas Dahlbæk Dec 11 '14 at 17:23

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