Let $X \subset \mathbb{R}$ and let $f,g : X\rightarrow X$ be continuous functions such that $f(X)\cap g(X) = \emptyset$ and $f(X)\cup g(X) = X$.

Which one of the following sets cannot be equal to $X$ ?

A. $[0, 1]$

B. $(0, 1)$

C. $[0, 1)$

D. $\mathbb{R}$

Please explain all options.

My approach:

Simply we can see the conditions imply that $X$ is disconnected. Continuity implies that $f([0,1])$ and $g([0,1])$ are compact. From the above conditions. Compact set would not be connected so $X\ne [0,1]$.


1 Answer 1


If $X$ were compact, so would $f[X]$ and $g[X]$ be, and they'd be two closed disjoint non-empty subsets of $X$ that form a partition of $X$, so then $X$ is not connected.

So we know a compact $X$ cannot be connected.

  • $\begingroup$ I know image of compact set is compact and [0,1] is compact and so f[0,1] and g[0,1] if they were disjoint then satisfies the condition. But option a) is false given $\endgroup$
    – Unknown
    Nov 17, 2018 at 6:33
  • $\begingroup$ Henno@ explain all options I have mentioned it. $\endgroup$
    – Unknown
    Nov 17, 2018 at 6:35
  • $\begingroup$ @JohnNash There is only one right answer I suppose, and now you've identified it. $\endgroup$ Nov 17, 2018 at 6:37
  • 1
    $\begingroup$ @JohnNash You know for sure that the option A is right: $X$ cannot be $[0,1]$, $X$ can be $\mathbb{R}$ or $[0,1)$ or $(0,1)$, but you have to come up with examples for that (just draw pictures). BCD are wrong. $\endgroup$ Nov 17, 2018 at 6:41
  • $\begingroup$ @JohnNash B and D are homeomorphic, so an example for one can be turned into an example for the other. For $[0,1)$ it's easy: $f(x)=\frac{1}{2} x$ and $g(x) = \frac{1}{2}(x+1)$ will do, I think. $\endgroup$ Nov 17, 2018 at 6:44

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