Let $A$ be a subset of $\mathbb{R}$, and let $f:A\to\mathbb{R}$ be a function. Now, let $\{r_{n}\}$ be any rational sequence in $A$, and let $\{s_{n}\}$ be any irrational sequence in $A$.
Suppose that $r_{n},s_{n}\to a\in A$ as $n\to\infty$, and that $f(r_{n}),f(s_{n})\to f(a)\in\mathbb{R}$ as $n\to\infty$.
Is it enough so that $f$ is continuous at $x=a$?
The above two sequences are not "any" sequence in $A$.
But, it seems to be the density of rational and irrational would give such weak condition to Sequential criterion for continuity.
It just my opinion. Is it possible?
Give some comments or counter-examples! Thank you!
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $x\in\{a_n\}\cup\{b_n\}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.