Continuity on open interval

Question

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

Answer 1

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?

Answer 2

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 $b$.

Answer 3

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.

Answer 4

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 $0$.