# Continuity on open interval

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 $(a,b)$.

• Continuity is a property of functions, not intervals. It doesn't make sense to say an interval is continuous or discontinuous. – Michael Albanese Aug 4 '13 at 20:26
• @MichaelAlbanese I can't understand what you mean to explain.can you please elaborate.. – spectraa Sep 19 '14 at 5:45

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?

• no, not quite, sorry. Is it because we'll always be able to find a read # that's less than b but more than a? I mean that, there will always be a continuous stream of numbers because the amount of numbers between a & b is infinite, is that why? thanks for your input...I'm only just starting with calculus. – Emi Matro Aug 4 '13 at 21:06
• @user4150 No prob, let's start at the beginning. Forget about $a$ and $b$ for the moment. Do you see why $f$ is continuous on $(0,1)$? – Chris Culter Aug 4 '13 at 21:10
• No, I don't really...I think I kind of understand the concept, but mathematical notation/symbols are a little confusing. Can you explain in words please – Emi Matro Aug 4 '13 at 21:14
• @user4150 Sure, it gives me an excuse to try out the chat feature! Please drop by at chat.stackexchange.com/rooms/9960/continuity-on-open-intervals – Chris Culter Aug 4 '13 at 21:19
• sorry, i need 20 rep to chat – Emi Matro Aug 4 '13 at 21:21

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 $b$.

• @PeterTamaroff I never said that the function was not defined at $a$ and $b$. – Daniel Robert-Nicoud Aug 4 '13 at 20:27
• My bad. The problem is you're misaddressing the OP's issue. – Pedro Tamaroff Aug 4 '13 at 20:28
• @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 4 '13 at 20:31

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.

• 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 4 '13 at 20:42
• What do you mean by "if $a$ is open"? – Michael Albanese Aug 4 '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 4 '13 at 20:49
• @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 4 '13 at 20:49
• @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 4 '13 at 20:53

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 $0$.