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

• Aug 14th 2009, 08:06 PM
numberstrong
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 $\displaystyle f,g :X \rightarrow Y$ both continuous and with $\displaystyle Y$ in the order topology, the set $\displaystyle \{x \ | \ f(x) \leq g(x) \}$ is closed.

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

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

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

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

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

Can anyone help with this? Thanks!

-Steve
• Aug 14th 2009, 09:28 PM
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.
• Aug 14th 2009, 10:32 PM
numberstrong
Quote:

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 $\displaystyle Y$ other than that it is a topological space with the order topology and nothing on $\displaystyle X$ other than that it is an arbitrary topological space.
• Aug 15th 2009, 02:27 AM
algtop
Quote:

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

(1) $\displaystyle f(x)$ is the immediate successor of $\displaystyle g(x)$. In this case, let $\displaystyle U \subset Y$ be defined by $\displaystyle U = \{y \ | \ y > g(x)\}$, and let $\displaystyle V \subset Y$ be defined by $\displaystyle V = \{y \ | \ y < f(x) \}$.
(2) There exists some $\displaystyle y_0$ satisfying $\displaystyle f(x_0) > y_0 > g(x_0)$ for an arbitrary x_0 in X; in this case, define $\displaystyle U \subset Y$ as $\displaystyle U = \{y \ | \ y > y_0\}$, and define $\displaystyle V \subset Y$ as $\displaystyle V = \{y \ | \ y < y_0\}$ for some such $\displaystyle y_0$.

For 2, let $\displaystyle x_0 \in A^c$. Then, $\displaystyle f^{-1}(U) \cap g^{-1}(V)$ is an open set containing $\displaystyle x_0$ (f and g are continous and an intersection of open sets is an open set) and is contained in $\displaystyle A^c$. Since $\displaystyle x_0$ is an arbitrary point $\displaystyle A^c = \{x \in X \ | \ f(x) > g(x) \}$ and the choice of $\displaystyle y_0$ is followed by the choice of $\displaystyle x_0$, we conclude that $\displaystyle A^c$ is open.

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

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