  Continuity [10/13.1] Let (X, d ) and (Y, d 1 ) be metric spaces. Let f : X &#174; Y be a constant function.
That is &#36; a y 0 &#206; Y &#39; for all x &#206; X, f(x) = y 0 . Show f is continuous. Since | f(x) - f(x o ) | = | y - y 0 | = 0 &#34; x 0 &#206; X, d 1 (f(x),f(x 0 )) = 0 e [10/13.2] The identity function f(x) = x is continuous because: | f(x) - f(x 0 ) | = | x - x 0 | d e [10/13.3] Let (R, d ), with R the real numbers and d(a,b) = | a - b |. Now, we define f by f : R &#174; R by: f(a) = {0 if a is rational, a if a is irrational} We show that f is not continuous at any point r , r &#185; 0. If r is irrational and x 0 &#206; Q, then | f(r) - f(x 0 ) | = | r - 0| = r . Thus, e 1 > r and | r - x 0 | d 1 . However, we know that there is an arbitrarily small distance between r and r 0 &#206; Q, with | f(r 0 ) - f(x 0 ) | = 0. But if | r 0 - x 0 | > | r - x 0 |, then e 2 > e 1 , but | r 0 - x 0 | > | r - x 0 | implies that d 1 d 2 , and this is a contradiction. Therefore, f is not continuous at any point r &#185; 0. 
