# Difference between closed and open intervals for continuous functions

Say I have a continuous function $f(x)$ on the open interval $(0, 1)$. There is no limit on how large $|f(x)|$ can be apart from the fact that it can't be infinity, it will always be a real number.

Now say I have a continuous function $f(x)$ on the closed interval $[0, 1]$. This too has no limit on how large $|f(x)|$ can be apart from the fact that it can't be infinity, it will always be a real number.

So it seems like there is no difference between them...and they are both unbounded? But from what I've read continuous functions on closed intervals are bounded...

This seems like a chicken and an egg scenario to me. Can anyone set me straight on the difference between continuous functions on open intervals vs. closed intervals?

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An open interval does not include its endpoints, and is indicated with parentheses. For example (0,1) means greater than 0 and less than 1. A closed interval includes its endpoints, and is denoted with square brackets. For example [0,1] means greater than or equal to 0 and less than or equal to 1.[source: en.wikipedia.org/wiki/Interval_%28mathematics%29] – Romantic Electron Jan 19 '15 at 4:38

Compare the statements:

1. For every $M>0$, there exists a function $f \in C([a,b])$ such that $\max_{x \in [a,b]} f(x)=M$;
2. For every $f \in C([a,b])$ and every $M>0$, there exists a point $x_M \in [a,b]$ such that $|f(x_M)|>M$.

If you fix the function $f\in C([a,b])$, then $f$ is counded by a constant (by Weierstrass' theorem). This is fals for $f \in C(a,b)$ (for instance $f(x)=(x-a)^{-1}$ is a counter-example). However, you cannot choose the constant independently of $f$.

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The discrepancy you see is accounted for by the following fact: if $f$ is unbounded on $(0,1)$ then it has to be either undefined or discontinuous at $0$ or $1$, and hence is not defined (or not continuous) on $[0,1]$.

Consider $f(x)=\dfrac{1}{x}$. This function is unbounded on $(0,1)$, but it is not defined on $[0,1]$ because it takes no value at $x=0$.

The proof of this fact is a fairly simple consequence of the result that any continuous function $f : [a,b] \to \mathbb{R}$ is bounded. If $f$ is bounded on $[a,b]$ then it is too on $(a,b)$. So if a continuous function is unbounded on $(a,b)$ then it must not be defined (or, at least, not continuous) on $[a,b]$, or else it would contradict this result.

Intuitively, a continuous function is allowed to misbehave at the endpoints of an open interval (because it doesn't have to be defined at the endpoints), but it must behave itself on a closed interval because closed intervals contain their endpoints.

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"Now say I have a continuous function f(x) on the closed interval [0,1]. This too has no limit on how large |f(x)| can be apart from the fact that it can't be infinity, it will always be a real number."

This is false. Since $[0, 1]$ is compact, $f(x)$ is bounded on $[0, 1]$.

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You pointed out the mistake in the question but that doesn't make it an answer.Please update it – Romantic Electron Jan 19 '15 at 4:40

Actually, there is a difference. The closed interval is a compact set. Hence the image of $f$ is compact too. Compact sets in $\mathbb{R}^k$ are closed and bounded per the Heine-Borel theorem. Since the image is bounded, it has upper and lower bounds. Since it's closed, $f$ must reach its upper and lower bounds in the interval.

For the open interval, however, the image of $f$ doesn't have to be bounded. $\tan(x)$ on $(-\pi/2, \pi/2)$ is an example.

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$f(x)$ is bounded in $[0,1]$ by Extreme value Theorem

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