Nowhere continuous function: Difference between revisions

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More general definitions of this kind of function can be obtained, by replacing the [[absolute value]] by the distance function in a [[metric space]], or the continuity definition by the definition of continuity in a [[topological space]].
 
One example of such a function is the [[indicator function]] of the [[rational number]]s, also known as the ''[[Dirichlet function]]''. This function is written ''I''<sub>'''Q'''</sub> and has [[___domain (mathematics)|___domain]] and [[codomain]] both equal to the [[real number]]s. ''I''<sub>'''Q'''</sub>(''x'') equals 1 if ''x'' is a [[rational number]] and 0 if ''x'' is not rational. If we look at this function in the vicinity of some number ''y'', there are two cases:
*If ''y'' is rational, then ''f''(''y'')&nbsp;=&nbsp;1. To show the function is not continuous at ''y'', we need to find an &epsilon; such that no matter how small we choose &delta;, there will be points ''z'' within &delta; of ''y'' such that ''f''(''z'') is not within &epsilon; of ''f''(''y'')&nbsp;=&nbsp;1. In fact, 1/2 is such an &epsilon;. Because the [[irrational number]]s are [[dense set|dense]] in the reals, no matter what &delta; we choose we can always find an irrational ''z'' within &delta; of ''y'', and ''f''(''z'')&nbsp;=&nbsp;0 is at least 1/2 away from 1.
*If ''y'' is irrational, then ''f''(''y'')&nbsp;=&nbsp;0. Again, we can take &epsilon;&nbsp;=&nbsp;1/2, and this time, because the rational numbers are dense in the reals, we can pick ''z'' to be a rational number as close to ''y'' as is required. Again, ''f''(''z'')&nbsp;=&nbsp;1 is more than 1/2 away from ''f''(''y'')&nbsp;=&nbsp;0.