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
Best Answer
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 $b$.