$f:[0,1] \to RR$...

[TomSawyer]

Determine all values of $a \in RR$ such that there exists a function $f:[0,1] \to RR$ fullfilling the following inequality for all $x\ne y$: $|f(x)-f(y)|\ge a$.

[Thomas]

I guess only a=0 can satisfy this condition... isn't it?

[Camillo]

That was also my idea which means $f(x) = k $ with $k in RR $ ,but I have no idea how to show it.

[Thomas]

If the answer is correct, I can suggest a "cardinality" argument... but I'm not sure.... you can try

[Camillo]

No conditions are indicated for the function $ f $ ; so it could also be discontinuous in any point of its domain .
Does this help to find another solution in addition to $a=0$ ?

[TomSawyer]

Those conditions are fullfilled for all $a\le0$, of course . To prove it, consider that $[0,1]$ is not a countable set, as Thomas suggested.

[Luke1984]

Let be $a>0$.
Suppose exists a function $f:[0,1]->RR$ such that $|f(x)-f(y)| \geq a$ for every $x ne y$.
Let be $B:=\{I\in P(RR): $ exists $k\in ZZ$ such that $ I=[ka,(k+1)a) \}$.
For every $x\in [0,1]$ there is $I_x in B$ such that $x in I_x$ (because the collection $B$ covers $RR$).
Moreover $I_x$ is unique, because the elements of $B$ are disjoint.
Finally if $x,y in [0,1]$, we have $I_x ne I_y$ (if not we would have $|f(x)-f(y)| < a$).
So there is an injection $g$ from $[0,1]$ to $B$.
On the other hand an injection from $B$ to $ZZ$ obviously exists.
So there is an injection from $[0,1]$ to $ZZ$, and this is absurd.

[TomSawyer]

Ok, that's exactly the proof I know. Cheers.