<p>The function \(f(x) = 1\), if \(x\) is rational<br>\(= 0\), if \(x\) is irrational<br>is discontinuous at all points \(x\).</p><p><em>State whether this statement is true or false.</em></p>
Step-by-Step Solution
Key Concept: A function is continuous at a point if the limit exists and equals the function value; for Dirichlet's function, any neighborhood of any point contains both rationals and irrationals, so the limit doesn't exist at any point.
<p><strong>Step 1:</strong> Recall that f(x) is continuous at x = a iff <span style='color:blue'>lim<sub>x→a</sub> f(x) = f(a)</span></p><p><strong>Step 2:</strong> For ANY point a ∈ ℝ, consider lim<sub>x→a</sub> f(x). In every δ-neighborhood of a:</p><ul><li>Rationals exist with f(x) = 1</li><li>Irrationals exist with f(x) = 0</li></ul><p><strong>Step 3:</strong> Since both values 0 and 1 appear arbitrarily close to a, the limit <span style='color:red'>does not exist</span> at any point a.</p><p><strong>Step 4:</strong> Since lim<sub>x→a</sub> f(x) doesn't exist for any a ∈ ℝ, the function cannot be continuous at any point.</p><p><strong>Step 5:</strong> We can verify: even if a is rational (f(a)=1), we can find irrationals arbitrarily close where f = 0 ≠ 1 = f(a), violating the ε-δ definition.</p><p>∴ <strong>Answer: TRUE</strong> — The function is discontinuous at ALL points.</p>
Correct Answer: A