give the calculation step by step as like as full answer in 2 steps and explanation block in every steps.
And final answer summary
Definition of Parameters
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give the calculation step by step as like as full answer in 2 steps and explanation block in every steps. 
And final answer summary
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VIPSolutions · Accepted Answer

**Step 1: Prove continuity of $ h(x) $ at every irrational number $ x_0 $** **Explanation:** To show continuity at an irrational $ x_0 $, use the definition of continuity: For every $ \epsilon > 0 $, there exists $ \delta > 0 $ such that for all $ x $ with $ |x - x_0| < \delta $, it follows that $ |h(x) - h(x_0)| < \epsilon $. Since $ x_0 $ is irrational, $ h(x_0) = 0 $. Consider any $ \epsilon > 0 $. Choose a positive integer $ N $ so that $ 1/N < \epsilon $. In any interval around $ x_0 $, there are only finitely many rational numbers of the form $ m/n $ with $ n \leq N $. Make the interval small enough so that all rational numbers in it have denominators $ n > N $, which ensures $ h(x) = 1/n < 1/N < \epsilon $ for any rational $ x $ in the interval. For irrational $ x $ in the interval, $ h(x) = 0 $. Thus, for all $ x $ in this neighborhood, $ |h(x) - 0| < \epsilon $. --- **Step 2: Prove discontinuity of $ h(x) $ at every rational number $ x_0 $** **Explanation:** Let $ x_0 = m_0/n_0 $ (in lowest terms, $ n_0 > 0 $), so $ h(x_0) = 1/n_0 $. To show discontinuity, use the definition: A function is discontinuous at $ x_0 $ if there exists $ \epsilon_0 > 0 $ such that for every $ \delta > 0 $, there is some $ x $ with $ |x - x_0| < \delta $ and $ |h(x) - h(x_0)| \geq \epsilon_0 $. Pick $ \epsilon_0 = 1/(2n_0) $. For any $ \delta > 0 $, there exists an irrational $ x $ in $ (x_0 - \delta, x_0 + \delta) $ (irrationals are dense). For such $ x $, $ h(x) = 0 $. Thus, $$ |h(x) - h(x_0)| = |0 - 1/n_0| = 1/n_0 > 1/(2n_0) = \epsilon_0 $$ So $ h $ is not continuous at $ x_0 $. --- **Final Answer Summary:** Thomas's function $ h(x) $ is continuous at every irrational number because for any small neighborhood around an irrational, the values of $ h(x) $ can be made arbitrarily close to 0. However, $ h(x) $ is discontinuous at every rational number because any neighborhood around a rational contains irrationals where the function value jumps by a fixed amount.

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