f(x) 不连续, f(f(x))能连续吗？

yma16 2024-03-21 07:05:49 ( reads)

以下是人工智能的回答，好像不太清楚。请看看。

Certainly! Here's an example:

Let's define a function $f (x)$ as follows:

x & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \] This function is discontinuous at every irrational number because \( f(x) = 0 \) for irrational \( x \), but \( f(x) = x \) for rational \( x \), causing a jump discontinuity. Now, let's consider \( f(f(x)) \): \[ f(f(x)) = f\left(f(x)\right) = f\left(\begin{cases} x & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases}\right) \] If \( x \) is rational, \( f(x) = x \), so \( f(f(x)) = f(x) = x \), which is continuous. If \( x \) is irrational, \( f(x) = 0 \), so \( f(f(x)) = f(0) = 0 \), which is also continuous. Thus, \( f(f(x)) \) is continuous everywhere, despite \( f(x) \) being discontinuous.

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