Originally Posted by

**numberstrong** I am trying to find out whether or not this is true.

**Let $\displaystyle X$ be an arbitrary topological space, and let $\displaystyle Y$ have the order topology. If $\displaystyle f,g: X \rightarrow Y$ are continuous, and if $\displaystyle f(x_0) > g(x_0)$ for a specific $\displaystyle x_0 \in X$, then there exists an open set $\displaystyle U \subset X$, containing $\displaystyle x_0$, such that $\displaystyle f(x) > g(x)$ for all $\displaystyle x \in U$.**

If I can show this, then my question in the other thread becomes easier.

So far, my intuition tells me that the best way to proceed would be to use the definition of continuity that says $\displaystyle f:X \rightarrow Y$ is continuous at a point $\displaystyle x_0 \in X$ if for each open set $\displaystyle V \subset Y$ containing $\displaystyle f(x_0)$, we can find an open set $\displaystyle U \subset X$, containing $\displaystyle x_0$, such that $\displaystyle f(U) \subset V$. I haven't gotten much further than that, though.

-Steve