Interactive Real Analysis - part of MathCS.org

• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 6.1. Limits
• 6.2. Continuous Functions
• 6.3. Discontinuous Functions
• 6.4. Topology and Continuity
• 6.5. Differentiable Functions
• 6.6. A Function Primer
• 7. The Integral
• 8. Sequences of Functions
• 9. Historical Tidbits
• Java Tools

Example 6.2.8(b):

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The function f(x) = x² is continuous on [0, 1]. Is it uniformly continuous there ?

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This function (as you could guess from its graph) is uniformly continuous on the closed interval [0, 1]. To prove it, note that

| f(t) - f(s) | = | t - s | | t + s | < | t - s | 2

because s and t are in the interval [0, 1]. Hence, given any > 0 we can simply choose = / 10 (or something similar) to prove uniform convergence. Can you fill in the details ? A similar argument, incidentally, would work on the interval [0, N] for any number N, but it would fail for the interval [0, ). So, if this function then uniformly continuous on the interval [0, ) ? That's the next example.