f(x) 不连续, f(f(x))能连续吗？

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.

所有跟帖： 

• 稍微改一下就好了吧？f(x)=0 for all irrational x, =1 for the rest; then -longsky- ♂ (21 bytes) () 03/21/2024 postreply 09:34:48

• 你的清楚。谢谢。 -yma16- ♂ (0 bytes) () 03/21/2024 postreply 16:09:24