Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$. Show that $\min\{x_n, y_n\} \rightarrow \min\{x, y\}$

Question

Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$. Show that $\min\{x_n, y_n\} \rightarrow \min\{x, y\}$

I'm not really sure how to show this. If $x_n$ converges to $x$ and $y_n$ converges to $y$, then wouldn't it already trivially follow that $\min\{x_n, y_n\}$ converges to $\min\{x, y\}$?

I know that I'm missing something, and any hints would be appreciated.

Answer 1

You just need a nice statement about the min, and you are done.

Lemma : $\min\{ a,b\} = \dfrac{a+b - |a-b|}{2}$.

Proof : If $a \geq b$, then $|a-b| = a-b$, so that $a+b-a+b = 2b$. Otherwise, $a < b$, so $|a-b| = b-a$, and we have $a+b + a - b = 2a$. After division by $2$, we see this corresponds to the minimum function.

Note that $|\min\{x_n,y_n\} -\min\{x,y\}|= \left|\frac{(x_n + y_n - |x_n - y_n|) - (x + y - |x-y|)}2 \right|$.

We use triangle inequality: $$ \left|\frac{(x_n + y_n - |x_n - y_n|) - (x + y - |x-y|)}2 \right| \leq \left|\frac{x_n-x}2\right| + \left|\frac{y_n-y}2\right| + \left|\frac{|x_n-y_n| - |x-y|}2\right| $$

Observe the terms on the right hand side. The first can be made small, since $x_n \to x$. The second can be made small, since $y_n \to y$. The third canbe made small, since $x_n - y_n \to x-y$.

Hence, we have convergence.

Inductively, I encourage you to show this corollary:

For any finite of number of convergent sequences $s_1...s_n$, $\min_{1 \leq i \leq n}\{s_i\} $ will converge to the minimum of the limits.

Alternately, you can also show this for maximum, because for the maximum of two quantities, there is a similar formula: $$ \max\{a,b\} = \frac{a + b + |a-b|}{2} $$

Hence, the same argument will follow.

Answer 2

We have that $$ \min\{x_n,y_n\}=\frac{x_n+y_n-|x_y-y_n|}2. $$ Using the properties of limits and the continuity of the absolute value, we get $$ \frac{x_n+y_n-|x_y-y_n|}2\to\frac{x+y-|x-y|}2=\min\{x,y\} $$ as $n\to\infty$.