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Let $X⊂\mathbb R$ and let $f,g:X→X$ be a continuous functions such that $f(X)∩ g(X)=∅$ and $f(X)∪g(X)=X$. Which one of the following sets cannot be equal to $X$?

  • $[0,1]$
  • $(0,1)$
  • $[0,1)$
  • $\mathbb R$

I think its $\mathbb R$, as $X⊂\mathbb R$ but I am not sure of how other options will behave here.

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    If I'm not mistaken, all of these are connected spaces. The continuous image of a connected space is again connected. – 2017-01-14
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    @emka I've been hung up on that too. I may also be mistaken, but I think that although $f, g$ are continuous, they are not necessarily open (that is, they need not send open sets to open sets). If they aren't, then $X$ isn't necessarily disconnected. – 2017-01-14
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    I see. This may throw a wrench into what I wrote as an answer. – 2017-01-14
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    @Jon Garrick, from where do you get the question? – 2017-11-12

3 Answers 3

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Suppose $X=[0,1]$; $f(X)$ and $g(X)$ are compact intervals since the image of a compact set by a continuous map is compact and the image of a connected set by a continuous map is connected. Write $f(X)=[a,b]$, $g(X)=[c,d]$ you may suppose $0\in f(X)$ so $f(X)=[0,b]$. $c>b$ otherwise, $[0,b]\cap [c,d]$ is not empty. This implies that $b+(c-b)/2$ is not in $f(X)\bigcup g(X)$. Contradiction.

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I'm open to discussion if this is incorrect:

Let $X=(0,1)$. Then $f(X)$ and $g(X)$ are both open sets. Now suppose that $f(X) \cap g(X)=\emptyset$ and $f(X)\cup g(X)=X$.

Let $a=\sup\{(0,1)\setminus g(X)\}$ and $b=\inf\{(0,1)\setminus f(X)\}.$ Then $a

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    You are wrong!!! Answer is $[0,1]$ – 2017-11-12
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Since $[0,1]$ is compact, also $f([0,1])$ and $g([0,1])$ are compact and thus closed.

Let $[0,1]$ be the disjoint union of $f([0,1])$ and $g([0,1])$. Because $f([0,1])=[0,1] \setminus g([0,1])$ the set $f([0,1])$ must be open. This contradicts $f([0,1])$ being closed because the only sets in $[0,1]$ which are both open and closed are $[0,1]$ and $\emptyset$. Because $f([0,1])$ and $g([0,1])$ have at least one element, $f([0,1])$ cannot be $[0,1]$ or empty.