The meaning of $\min()$ in an inequality I am working on a problem but I am stuck because I do not know how to interpret this part of the problem:
$\varepsilon$ is a positive real number. $x, x_0,$ and $y_0$ are real numbers.
$$ |x-x_{0}|<\min \left(\frac{ \varepsilon  }{2(|y_{0}|+1)},1\right) $$
The rest of the problem includes another inequality to prove a third inequality. I am stuck because I do not know how to interpret the "$\min ( )$" part of this inequality. This is for a first year calculus course - second week.
 A: The min means take the minimum of the arguments.  Generally, $\epsilon$ is expected to be quite a small positive number.  Proofs that there is an acceptable delta sometimes fail when $\epsilon$ gets large.  So we calculate an acceptable delta assuming $\epsilon$ is small (which depends upon $\epsilon$), then put a hard upper limit (here $1$) to handle the case if $\epsilon$ gets large.
A: The other answers have explained what it is being used for here; I'd like to explain just $\min$.
$\min$ is a function, which happens to be part of an inequality here. You could write $\min$ as a piecewise function definition:
$$
\min(a,b) = \begin{cases}
a &\text{if } a ≤ b \\
b &\text{if } b ≤ a
\end{cases}
$$
It is also sometimes written $\min\{a, b\}$, meaning to take the minimum element of the set $\{a, b\}$ (which could have more or less than 2 elements).
A: Since $\varepsilon$ is arbitrary it is possible that $\frac{\varepsilon}{2(|y_0|+1)}>1$, while the proof wants to assert $|x-x_0|$ is small.
Therefore this reads as $|x-x_0|$ is less than the minimal number between $1$ and $\frac{\varepsilon}{2(|y_0|+1)}>1$. In a sense this is a "cutoff" to ensure the value is at most $1$.
A: Probably the simplest way is to interpret it as two inequalities: $|x - x_0| < 1$ and $|x-x_0| < \frac{ \varepsilon  }{2(|y_{0}|+1)}$.  If this is an assumption, then you can use either inequality as needed.  If it is what you are trying to prove, then just prove both inequalities.
