Do discontinuous preferences imply no continuous utility function? I am trying to think of a preference relation that can be represented by a utility function but such that there does not exist a continuous utility function. 

I know that you can represent continuous preferences with a discontinuous utility function, but I am not sure if the opposite is true. I am struggling to show that NO such continuous utility function exists.

My thoughts are that if you can define something with discontinuous preferences then maybe you can use this to imply that there do not exist any continuous utility functions. Note that Lexiographic preferences will not work because I am interested in a preference that can be represented by a utility function (albeit discontinuous).
The easiest way to prove it is using the 'old' definition of continuity. 
$\succ$ is continuous iff whenever $x\succ y$, there exists neighborhoods of $x$ and $y$, $B_x, B_y$, such that all $z\in B(x)$ and $z'\in B(y)$, $z\succ z'$.

Suppose $x\succ y$. Because $u$ represents $\succ$, $u(x)&gt;u(y)$. Let $2\epsilon=u(x)-u(y)$. Because $u$ is continuous, there exists some $\delta&gt;0$ such that for all $z\in B_{\delta}(x)$, $u(z)&gt;u(x)-\epsilon$. Similarly, for all $z'\in B_{\delta}(y)$, $u(z')&gt;u(y)+\epsilon$. But then for all $z\in B_{\delta}(x)$ and $z'\in B_{\delta}(y)$, $z\succ z'$ as required.