Can there be two distinct, continuous functions that are equal at all rationals?

Akhil showed that the Cardinality of set of real continuous functions is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.

This isn’t an obvious step, so why is it true?

Answer

Without resorting to ε-δ arguments: Let $f$ and $g$ be continuous real functions and $f(x) = g(x)$ for all rational $x$. For any real number $c$ (in particular, an irrational $c$), there exists a Cauchy sequence of rational numbers such that $\lim_{n \to \infty}x_{n}=c$. Since $f$ and $g$ are continuous, $\lim_{n \to \infty}f({x_{n}})=f({c})$ and $\lim_{n \to \infty}g({x_{n}})=g({c})$. Since $x_n$ is rational, $f(x_n) = g(x_n)$ for all $n$, so the two limits must be equal and so $f(c) = g(c)$ for all real $c$.

Attribution
Source : Link , Question Author : Charles Stewart , Answer Author : zar

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