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For which subspace $X\subseteq \mathbb{R}$ with the usual topology and with$\{0,1\} \subseteq X$ will a continious function $f : X \rightarrow \{0,1\}$ satisfying $f(0) =0$ and $f(1) =1$ exist ?

$a)$$ X=[0,1]$

$b)$$X=[-1,1]$

$c)$$X=\mathbb{R}$

$d)$$[0,1] ⊄X$

i was thinking about the function $f(x) = x$ that is $f(0) =0$ and $f(1) =1$ and i don't know how to tackle this question

Any hints/solution

thanks u

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    $\begingroup$ $f\colon X \to \{0,1\}, x\mapsto x$ is not well defined on the four choices of $X$. For example, in a) $f(\frac12)=\frac12 \notin \{0,1\}$. $\endgroup$ – Babelfish Oct 1 '18 at 12:01
  • $\begingroup$ @Babelfish thanks u $\endgroup$ – jasmine Oct 1 '18 at 12:14
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Hint: the image of a connected set by a continuous function is again connected.

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  • $\begingroup$ thanks longa... $\endgroup$ – jasmine Oct 1 '18 at 12:15

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