Nowhere continuous function: Difference between revisions

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AIn [[[mathematics]], a '''nowhere continuous''' [[function]] is (tautologically) a function that is not [[continuous]] at any point of its [[function ___domain|___domain]]. Suppose Thatthat ''f'' is a function from [[real number]]s to sayreal numbers. Then, <i>f(x)</i> is nowhere continuous for each point <i>x</i> there is an <i>&epsilon; >0</i> such that for each <i>&delta; >0</i> we can find a point <i>y</i> such that <i>|x-y|<&delta; </i> and <i>|f(x)-f(y)|>&epsilon; </i>. The Basically,import of this statement is a statement that at each point we can choose a distance such that points arbitrarily close to our original point are taken at least that distance away from the function's value.
 
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 entire continuity definition by the definition of continuity in a [[topological space]].
 
OnOne example of such a function is a function <i>f</i> on the [[real number|real numbers]] such that <i>f(x)</i> is 1 if <i>x</i> is a [[rational number]], but 0 if <i>x</i> is not rational. If we look at this function in the vincinityvicinity of some number <i>y</i>, there are two cases:
 
If y is rational, then <i>f(y)</i>=1. To show the function is not continuous at y, we need find a single <i>&epsilon;</i> which works in the above definition. In fact, 1/2 is such an <i>&epsilon;</i>, since we can find an irrational number <i>z</i>arbitrarily close to y and <i>f(z)</i>=0, at least 1/2 away from 1. If y is irrational, then <i>f(y)</i>=0. Again, we can take <i>&epsilon;</i>=1/2, and this time we pick <i>z</i> to be an rational number as close to <i>y</i> as is required. Again, <i>f(z)</i> is more than 1/2 away from <i>f(y)</i>