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Per sagemath.org:

A function is said to be continuous on an interval when the function is defined at every point on that interval and undergoes no interruptions, jumps, or breaks. If some function f(x) satisfies these criteria from x=a to x=b, for example, we say that f(x) is continuous on the interval [a, b]. The brackets mean that the interval is closed -- that it includes the endpoints a and b. In other words, that the interval is defined as a ≤ x ≤ b. An open interval (a, b), on the other hand, would not include endpoints a and b, and would be defined as a < x < b.

So, if a function is continuous on an open interval $(a,b)$ or even semi-open such as $[a,b]$, $(a,b]$ is it bounded?

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No. Consider the function $f:(0,1)\to\mathbb{R}$ given by $f(x)=\frac{1}{x-1}$. Clearly $f(x)\to \infty$ for $x\to 1$, so it is not bounded. The same counterexample works on half-open intervals. However, if the interval is closed, continous implies bounded. This is one of the important theorems when starting to learn analysis.

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