# Thread: Need to show {x | f(x) ≤ g(x)} closed if f,g continuous.

1. ## Need to show {x | f(x) ≤ g(x)} closed if f,g continuous.

I'm sure this is pretty elementary but I am intellectually needy.

I need to show that with $f,g :X \rightarrow Y$ both continuous and with $Y$ in the order topology, the set $\{x \ | \ f(x) \leq g(x) \}$ is closed.

My work so far: I am trying to show that the complement is open. If we denote the given set by $A$, then the complement is $A^c = \{x \ | \ f(x) > g(x) \}$. For a particular $x$ satisfying $f(x) > g(x)$ we have one of two cases:

(1) $f(x)$ is the immediate successor of $g(x)$. In this case, let $U \subset Y$ be defined by $U = \{y \ | \ y > g(x)\}$, and let $V \subset Y$ be defined by $V = \{y \ | \ y < f(x) \}$.

(2) There exists some $y_0$ satisfying $f(x) > y_0 > g(x)$; in this case, define $U \subset Y$ as $U = \{y \ | \ y > y_0\}$, and define $V \subset Y$ as $V = \{y \ | \ y < y_0\}$ for some such $y_0$.

In both of these cases, $U,V$ are open sets (since $Y$ has the order topology), and since $f,g$ are continuous, we have that $f^{-1}(U)$ and $g^{-1}(V)$ are both open in $X$.

That's the work that I've done so far, with some coaching from a professor. Now what I would like to do is create a collection of sets $\{U_x\}_{x \in X}$, and then claim that since $f^{-1}(U_x)$ is open for each $x$ ( $U_x$ is open and $f$ is continuous), we can take the union $\bigcup_{x \in X} U_x$ and claim it to be open in $X$. This is where my reasoning gets a little blurry, though. Is this union really the same as $\{x \ | \ f(x) > g(x)\}$? It seems in a certain sense that it is, since each $f^{-1}(U)$ produces the set of inputs such that the outputs of $f(x)$ are greater than the outputs of $g(x)$ for that particular $x$. But in another sense it seems obvious that this does NOT compose the set in question, since if we allow $x$ to range over all possible values then the union will include a great many values for which in general it is not true that $f(x) > g(x)$. ALSO, and perhaps more vexingly, the current argument does not make use of the continuity of $g$.

Can anyone help with this? Thanks!

-Steve

2. if f(x0)>g(x0) for certain x0,let u=(f(x0)-g(x0))/3
then there exist v such that for any |x-x0|<v, |f(x)-f(x0)|<u,|g(x)-g(x0)|<u because of the continuuity.
Thus f(x)>g(x) for any |x-x0|<v
So it will be an open set.

3. Originally Posted by ynj
if f(x0)>g(x0) for certain x0,let u=(f(x0)-g(x0))/3
then there exist v such that for any |x-x0|<v, |f(x)-f(x0)|<u,|g(x)-g(x0)|<u because of the continuuity.
Thus f(x)>g(x) for any |x-x0|<v
So it will be an open set.

The problem does not state that the functions in question are real-valued, and so we cannot use things like "division by 3" in the answer. If they were both real-valued then the problem would be very easy. But we have no measures, no algebraic structure, no binary operations or any other structure on $Y$ other than that it is a topological space with the order topology and nothing on $X$ other than that it is an arbitrary topological space.

4. Originally Posted by numberstrong
$A^c = \{x \ | \ f(x) > g(x) \}$. For a particular $x$ satisfying $f(x) > g(x)$ we have one of two cases:

(1) $f(x)$ is the immediate successor of $g(x)$. In this case, let $U \subset Y$ be defined by $U = \{y \ | \ y > g(x)\}$, and let $V \subset Y$ be defined by $V = \{y \ | \ y < f(x) \}$.
(2) There exists some $y_0$ satisfying $f(x_0) > y_0 > g(x_0)$ for an arbitrary x_0 in X; in this case, define $U \subset Y$ as $U = \{y \ | \ y > y_0\}$, and define $V \subset Y$ as $V = \{y \ | \ y < y_0\}$ for some such $y_0$.
For 2, let $x_0 \in A^c$. Then, $f^{-1}(U) \cap g^{-1}(V)$ is an open set containing $x_0$ (f and g are continous and an intersection of open sets is an open set) and is contained in $A^c$. Since $x_0$ is an arbitrary point $A^c = \{x \in X \ | \ f(x) > g(x) \}$ and the choice of $y_0$ is followed by the choice of $x_0$, we conclude that $A^c$ is open.

For 1, let $x_0 \in A^c$; let $U = \{y \in Y \ | \ y > g(x_0)\}$ and $V = \{y \in Y \ | \ y < f(x_0) \}$. Again, $f^{-1}(U) \cap g^{-1}(V)$ is an open set containing $x_0$ and is contained in $A^c$ (A choice of $x_0$ is arbitrary in $A^c = \{x \in X \ | \ f(x) > g(x) \}$ as well ). Thus, $A^c$ is open.

Since $A^c$ is open, we conclude that A is closed.