# A Surjective Function from $[0,1] \setminus \{0.5\}$ to $[0,1]$

I'm taking Discrete Mathematics this semester and I got this question in one of my homework tasks. I've tried thinking about the solution over and over but can't seem to come up with anything. The question goes like this:

Is there a surjective function from $[0,1] \setminus \{0.5\}$ to $[0,1]$ such that $f(a)>f(b)$ implies $a>b$?

I must mention that sadly I cannot use any arguments involving cardinality.

• Hint: Intuitively you might come to the conclusion that such a function $f$ has to "miss" at least one value in $[0,1]$. How would you characterize this value in terms of $f$? – Did Dec 5 '16 at 13:07
• Assume such a function exists. Surjectivity implies that at least one argument maps onto $\frac{1}{2}$. What can you say about that argument? – R.G. Dec 5 '16 at 13:20
• The correct statement of the exercise is: "...such that $a<b$ implies $f(a)<f(b)$". Please check your notes. – Did Dec 5 '16 at 14:05
• @R.G. Not much... :-) – Did Dec 5 '16 at 14:05
• @Did Hey! Sorry for my English :) it's written like this: F(a)>F(b) -> a>b – user396201 Dec 5 '16 at 14:07

How about you map all values $x \in [0,0.1]$ to $10x$, and all others to 1?