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Given countinous function $f : [0; 1] \cup (2;3) \rightarrow \mathbb{R}$. I have to determine whether following sets can be images of the function. a) $[0;1]$ b) $[0; 1]\cup [2;3]$ c) $\{1, 2, 3\}$. I think that a)yes b) yes c) no. Am I right?

Another function and sets: $f : [0; 1] \cup [2;3] \rightarrow \mathbb{R}$ a) $[0;1]$ b) $[0; 1) \cup (2; 3]$ c)$[0;2]\cup (1;3]$. Now b) and c) seems to be tricky. What are the answers in this cases?

alex
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  • Yes, you are right. – MooS Feb 03 '15 at 16:36
  • Yes, you are right, but to get full score you need to, in case a and b, supply example functions that do in fact have those sets as images. In case c you need to give a good reason why it cannot be done. – Arthur Feb 03 '15 at 16:44

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