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If $f(x)$ is continuous at $x_0$, it is locally bounded at $x_0$.

My attempt at a solution: If $f(x)$ is continuous at $x_0$, then it is defined for $x \approx_{\delta} x_0$ for some $\delta >0$, and given $\epsilon > 0$, $f(x) \approx_{\epsilon} f(x_0)$ for $x \approx_{\delta} x_0$. Continuity is a local property, so we check that $f(x) \approx_{\epsilon} f(x_0)$ for each point $x_0 \in I$.

Here is where I get stuck since I am not too sure about how to proceed checking points in $I$. The definition of locally bounded says that there is an open interval $I$ such that for all $x_0 \in I$, $f(x)$ is bounded for $x \approx x_0$. So how does one show this boundedness?

Any help/suggestions would be greatly appreciated.

I am using the textbook Introduction to Analysis by Arthur Mattuck.

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Perhaps there is some confusion in the definition of terms. To be precise, a function $f$ is said to be locally bounded at a point $x_{0}$ if there is an open interval $I$ containing $x_{0}$ such that $f$ is bounded on $I$. This means there is a number $K > 0$ for which $|f(x)| < K$ for all $x \in I$.

Hint: For your problem you need to apply the definition of continuity at $x_{0}$ and choose $\epsilon = 1$. Then you should be able to get an interval $I$ containing $x_{0}$ in which $f$ is bounded.

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  • $\begingroup$ Thank you very much @Paramanand Singh, this is very helpful! $\endgroup$ – Jamil_V Oct 16 '13 at 18:02

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