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Does a horizontal line count as bounded? Also, how is $y=2^x$ bounded?

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Don't forget (as I did at first reading) that "bounded" can mean "bounded below" (as is the case for $y = 2^x$). – Austin Mohr Sep 11 '12 at 0:33
A horizontal line is the graph of a bounded function: there are values that $y$ will never exceed and values it will never fall below. However, a horizontal line is not a bounded set: there is no distance that is not exceeded by that between some pair of points on the line. – Michael Hardy Sep 11 '12 at 1:08
Bounded usually does not mean "bounded either above or below". That's why we have the separate phrases "bounded", "bounded below" and "bounded above". We usually say that a function is bounded if and only if it is bounded above and bounded below. – Euler....IS_ALIVE Sep 11 '12 at 4:59

That depends on what you mean by bounded. Typically we mean that its value doesn't increase (or decrease) past some finite number, but whether or not this is true depends on the interval we're considering. Both examples are bounded on any finite interval, or an interval $(-\infty,a)$ for any finite $a$. Only the horizontal line is bounded on intervals of the form $(a,\infty)$ or $(-\infty,\infty)$, since $2^x$ is left to get arbitrarily large in that instance.

Only a function with a vertical asymptote (such as $\frac 1 x$) can be unbounded over a finite interval.

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A function has a horizontal asymptote if $\displaystyle\lim_{\pm\infty} f=c$

This does not mean $f(x)\geq c$ or $f(x)\leq c$ for all $x$. So having an asymptote does not imply the function is bounded. Simple example:


(horizontal asymptote $y=0$, but $f(1)=1$ and $f(-1)=-1$)

$y=2^x$ is bounded because any basic exponential function cannot be negative: it is continuous over $\mathbb{R}$, strictly increasing, and $\displaystyle \lim_{x\to-\infty}2^x=0$. These conditions imply $2^x>0$ for all $x$. Therefore it is bounded, and it has an asymptote: $y=0$

Note: bounded does not imply horizontal asymptote either. Take $y=\sin x$, which is bounded (since $-1\leq\sin x\leq 1$), but does not have an asymptote.

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A horizontal line is bounded. If $y=c$, then $y$ is bounded by any number greater than or equal to $c$. Also, $y = 2^x$ is not bounded. Suppose it were bounded by $c$. Then taking $x > \frac{\log{c}}{\log{2}}$ gives a contradiction.

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A horizontal line $y=c$ is bounded above by $c$, since $y$ is never more than $c$, and also below by $c$, since $y$ is never less than $c$. In the case of $y=2^x$, $y$ does not have an upper bound (since we can always pick an $x$ to make $y$ as big as we want), but it is bounded below by $0$, since we can never choose $x$ to make $y$ less than or equal to $0$.

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$f(x)=2^x$ is not a bounded function since it doesn't have both upper and lower bounds on the values of $y$. But it has a lower bound: $y$ is always $\ge0$.

The arctangent function is an example of a function with two horizontal asymptote: $y=\arctan x\to\pm\pi/2\text{ as }x\to\pm\infty$, with "$+$" in both expressions or "$-$" in both. It is a bounded function since $-\pi/2\le y\le\pi/2$ no matter what $x$ is.

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