Question Number 1196 by 123456 last updated on 13/Jul/15
$$\mathrm{lets}\:{f}:\mathbb{R}\rightarrow\mathbb{Z},\:\mathrm{suppose}\:\mathrm{that}\:\exists{x}\in\mathbb{R}\:\mathrm{and} \\ $$$$\exists{y}\in\mathbb{R}\:\mathrm{with}\:{x}\neq{y}\:\mathrm{such}\:{f}\left({x}\right)\neq{f}\left({y}\right) \\ $$$$\mathrm{can}\:{f}\left({x}\right)\:\mathrm{be}\:\mathrm{conrinuous}? \\ $$
Answered by prakash jain last updated on 14/Jul/15
$${f}\left({x}\right)={a}\in\mathbb{Z} \\ $$$${f}\left({y}\right)={b}\in\mathbb{Z} \\ $$$$\mathrm{If}\:{f}\left(\right)\:\mathrm{is}\:\mathrm{continous}\:\mathrm{then}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{takes} \\ $$$$\mathrm{all}\:\mathrm{values}\:\mathrm{between}\:{a}\:\mathrm{and}\:{b}. \\ $$$$\mathrm{contradicts}\:{f}\left({x}\right)\in\mathbb{Z}. \\ $$