f,g continuous --> inequalities happen on open sets?

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August 15th 2009, 12:35 AM
numberstrong

f,g continuous --> inequalities happen on open sets?

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

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

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 $f:X \rightarrow Y$ is continuous at a point $x_0 \in X$ if for each open set $V \subset Y$ containing $f(x_0)$ , we can find an open set $U \subset X$ , containing $x_0$ , such that $f(U) \subset V$ . I haven't gotten much further than that, though.

-Steve
August 15th 2009, 02:35 AM
algtop

Quote:

Originally Posted by numberstrong

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

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

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 $f:X \rightarrow Y$ is continuous at a point $x_0 \in X$ if for each open set $V \subset Y$ containing $f(x_0)$ , we can find an open set $U \subset X$ , containing $x_0$ , such that $f(U) \subset V$ . I haven't gotten much further than that, though.

-Steve

Let X be an arbitrary topological space and A be a subset of X. For every x in A, if we have an open set U containing x and is contained in A, then A is open in X.

This is what I remeber and I think it is a key idea to attack your previous problem.
August 15th 2009, 02:42 AM
numberstrong

Quote:

Originally Posted by algtop

Let X be an arbitrary topological space and A be a subset of X. For every x in A, if we have an open set U containing x and is contained in A, then A is open in X.

This is what I remeber and I think it is a key idea to attack your previous problem.

The fact you state is one that I would use in the last steps of this proof; after showing that around each $x_0$ I could form an open $U \subset X$ such that $f(x)>g(x)$ for all $x \in U$ , I would then claim that the set $A = \{x \ | \ f(x) > g(x)\}$ must be open since around each of its members we can form an open set entirely contained within $A$ . But this fact does not help work out the intermediate steps in the problem.

What I need help with is showing that the beginning of the above paragraph is actually true: that around each $x_0$ I can form an open $U \subset X$ such that $f(x)>g(x)$ for all $x \in U$ .
August 15th 2009, 04:10 AM
algtop

Quote:

Originally Posted by numberstrong

What I need help with is showing that the beginning of the above paragraph is actually true: that around each $x_0$ I can form an open $U \subset X$ such that $f(x)>g(x)$ for all $x \in U$ .

Yes. This is a detailed description.

Let $h:X \rightarrow Y$ defined by h(x)=f(x)-g(x) for all x in X. We know that f and g are continuous on X. Thus, h is continuous on X.

Let $B = \{h(x) \in Y | h(x) > 0 \}$ . Let V be an open set containing a point y_0 in B and in contained in B. Since h is continuous, we have an open set U containing $x_0 \in X$ such that $h(U) \subset V$ . It follows that an arbitrary point in U, let's say x_1, satisfies that $h(x_1) \in V$ . Thus, $f(x)>g(x)$ for all $x \in U$ .
August 15th 2009, 04:17 AM
numberstrong

Quote:

Originally Posted by algtop

Let $h:X \rightarrow Y$ defined by h(x)=f(x)-g(x) for all x in X. We know that f and g are continuous on X. Thus, h is continuous on X.

There is no notion of subtraction in this problem; $Y$ is an ordered set, but does not have operations like addition and multiplication defined.
August 15th 2009, 05:19 AM
algtop

Quote:

Originally Posted by numberstrong

There is no notion of subtraction in this problem; $Y$ is an ordered set, but does not have operations like addition and multiplication defined.

Sorry, this is my studpid mistake introducing addition operation and 0.

I updated the previous post. See here
In the link, $f^{-1}(U) \cap g^{-1}(V)$ is the required open set containing x_0, satisfying f(x) > g(x) for all points x inside the open set $f^{-1}(U) \cap g^{-1}(V)$ .

So, I think your assumption is true.

Hope some other people verify this as well.