Sequence criteria and right-contiunity of a non-decreasing function We know that $f$ is right continous at $x$ iff for every sequence $(x_n)$ such that $x_i>x$ and $x_n \to x$ we have $f(x_n) \to f(x)$. My question is: Why does it suffice (in order to get a conculusion) for $f$ non-decreasing, to only check this property for decreasing sequences?
 A: Suppose that $\langle f(z_n):n\in\Bbb N\rangle$ converges to $f(x)$ whenever $\langle z_n:n\in\Bbb N\rangle$ is a strictly decreasing sequence converging to $x$. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence converging to $x$ such that $x_n>x$ for all $n\in\Bbb N$, and suppose that $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$.


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*Show that there are an $\epsilon>0$ and a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ of $\langle x_n:n\in\Bbb N\rangle$ such that $|f(x_{n_k})-f(x)|\ge\epsilon$ for each $k\in\Bbb N$.


To avoid getting the subscripts too deep, let $y_k=x_{n_k}$ for each $k\in\Bbb N$. Let $m_0=0$. Given $m_k$ for some $k\in\Bbb N$, there is an $m_{k+1}\in\Bbb N$ such that $m_{k+1}>m_k$ and $y_{n_{k+1}}<y_{n_k}$. (Why?) Then $\langle y_{n_k}:k\in\Bbb N\rangle$ is a strictly decreasing subsequence of $\langle y_k:k\in\Bbb N\rangle=\langle x_{n_k}:k\in\Bbb N\rangle$ converging to $x$.


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*Conclude that on the one hand $\langle f(y_{n_k}):k\in\Bbb N\rangle$ converges to $f(x)$, but on the other hand $|f(y_{n_k})-f(x)|\ge\epsilon$ for each $k\in\Bbb N$. 


This is clearly a contradiction, so no such sequence $\langle x_n:n\in\Bbb N\rangle$ can exist. Thus, if $f$ behaves nicely on strictly decreasing sequences converging to $x$, it must behave nicely on all sequences converging to $x$ from the right.
