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A function is said to be continuous on an open interval if and only if it is continuous at every point in this interval.

But an open interval $(a,b)$ doesn't contain $a$ and $b$, so we never actually reach $a$ or $b$, and therefore they're not defined, and points that are not defined are not continuous, in other words $f(a)$ and $f(b)$ don't exist which makes the interval $(a,b)$ discontinuous.

So what is this definition saying, because I thought that it can't be continuous at $a$ or $b$ since they are not defined (an open circle on the graph), but everywhere in between $a$ and $b$ it can still be continuous...

So is it just continuous between these points $a$ and $b$, and a jump discontinuity occurs at these two points? Why then does it say that it's continuous at every point in $(a,b)$, if we are not including $a$ and $b$?

Points on an open interval can be approached from both right and left, correct? why is it required to be continuous on open $(a,b)$ in order to be continuous on closed $[a,b]$, I don't understand this because $a$ and $b$ are not defined in $(a,b)$.

please help to understand

Emi Matro
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4 Answers4

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As you stated in the definition, $f:X\rightarrow Y$ is continuous on $(a,b)\subseteq X$ if it is continuous at every point of $(a,b)$. Since $a,b\notin(a,b)$, we can have a discontinuity there. For example the characteristic function of $(a,b)$, $\chi_{(a,b)}:\mathbb{R}\rightarrow\mathbb{R}$, is continuous in $(a,b)$ but discontinuous at $a$ and $b$.

Daniel Robert-Nicoud
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    @PeterTamaroff I never said that the function was not defined at $a$ and $b$. – Daniel Robert-Nicoud Aug 04 '13 at 20:27
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    My bad. The problem is you're misaddressing the OP's issue. – Pedro Aug 04 '13 at 20:28
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    @PeterTamaroff I think OP's issue is not quite clear. Daniel's answer may address it. The issue is something about what happens at the endpoints of the interval. Daniel has provided an example with a finite discontinuity. In my answer I have given an infinite discontinuity. Maybe the response will clarify matters. – Mark Bennet Aug 04 '13 at 20:31
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Think about the function $\frac 1x$ on the open interval $(0,1)$ - it is not defined at $0$, but this does not stop it being continuous on the interval - in fact it is continuous because the interval is open, and we never have to deal with the bad value $x=0$.

The function $\tan x$ for the interval $(-\frac{\pi}2,\frac {\pi}2)$ is continuous, with "problems" at both ends.

Perhaps you could explain your problem in relation to these functions, as it may help to tease out what your issue really is.

Mark Bennet
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  • I thought that $\lim_{x\rightarrow a} f(x)=f(a)$ means that $f(a)$ is defined and exists, how can it exist if $a$ is open and not defined? – Emi Matro Aug 04 '13 at 20:42
  • What do you mean by "if $a$ is open"? – Michael Albanese Aug 04 '13 at 20:43
  • @MichaelAlbanese the $a$ in $(a,b)$ is "open", I mean that it is not included as an endpoint, it is never reached, and thus it's not defined as an actual point – Emi Matro Aug 04 '13 at 20:49
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    @user4150 If the limit exists, and equals the value of the function (because the function is defined beyond the open interval on which it is known to be continuous) the function is then continuous on one side at $a$ - we know nothing from what you have said about the limit from the other side. It is possible for the limit to exist, but to be different from the value of the function at $a$ – Mark Bennet Aug 04 '13 at 20:49
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    @user4150: I understand what you mean, but it is an incorrect use of the word 'open'. Saying the $a$ in $(a, b)$ is open doesn't mean anything; you should say "$(a, b)$ is open". Part of understanding these ideas is getting your head around the terminology and its correct usage (in particular, to which objects it applies). It is especially important when using an online medium such as MSE. – Michael Albanese Aug 04 '13 at 20:53
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You might find it helpful to think about the fact that some functions that are continuous on $(a,b)$ can be extended to $[a,b]$ to give a function continuous on the interval. For instance, the function defined by $f(x) = 1$ for all $x \in (0,1)$ can be continuously extended to a function $g$ defined by $g(x) = 1$ for all $x \in [0,1]$. On the other hand, a function like $h(x) = \frac{1}{x}$ is continuous on $(0,1)$, but cannot be continuously extended at $0$.

Zach L.
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Let's consider a really simple function. That way, we can look at how the terminology is used without worrying about the function's peculiar behavior.

Let $f(x)=0$ for all $x$. Then I claim:

  • $f$ is continuous on the open interval $(0,1)$.
  • $f$ is also continuous on $(0,2)$.
  • $f$ is continuous on $(1,2)$.
  • $f$ is continuous on $(3,4)$.
  • $f$ is continuous on $(-2\pi,e)$.
  • For any real numbers $a<b$, $f$ is continuous on $(a,b)$.

These statements are all true! Do you see why?

Chris Culter
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