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Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous. Show that the set $A = \{x:f(x)\leq g(x)\}$ is closed in $X$.

I am completely stumped on this problem. As far as I can tell I've either got the task of proving $A$ gathers its limit points, or showing that there is a closed set $V \subset Y$ such that $f^{-1}(V)$ or $g^{-1}(V)$ is equal to A, which would prove $A$ closed since both $f$ and $g$ are continuous.

The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set $V$. The fact that I don't know if $f$ or $g$ are injective means I keep running into the problem of having $f^{-1}(V)$ or $g^{-1}(V)$ give me extra points not in $A$. And even if I were able to construct $V$ this wouldn't guarantee it was closed.

I suspect I may need to use the continuity of both $f$ and $g$ together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks.

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5 Answers 5

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To show the set $A$ is closed, we only need to proof

$A^c=\{x\in X:f(x)>g(x)\}$ is open.

Proof: For any point $x \in A^c$, $f(x)>g(x)$; and the order topological space is Hausdorff, then there exist disjoint open sets $U_1$ with $f(x) \in U_1$ and $U_2$ with $g(x) \in U_2$ in $X$, such that for any point $y \in U_1$ and $z \in U_2$, $f(y)>g(z)$. Let $U= f^{-1}(U_1) \cap g^{-1}(U_2)$ is an nonempty open set in $X$: for $x \in U$. Obviously, we see $x \in U \subset A^c$, which implies the set $A^c$ is open.

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    $\begingroup$ On line two of your proof I think you meant "in $Y$" not "in $X$". Also could you explain how it follows that in these two disjoint open sets in $Y$ we always have $f(y) > g(z)$ for any $y\in U_1$ and $z\in U_2$? I don't see how this follows from what you have above it. Thanks. $\endgroup$
    – Thoth
    Jul 2, 2012 at 17:33
  • $\begingroup$ It is from the special structer of the space $Y$ with ordinal topology, and I think it's not a difficult thing, therefore, I omitt the process of the proof:) $\endgroup$
    – Paul
    Jul 3, 2012 at 0:18
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    $\begingroup$ so arrogant...You must understand that the questioner is a first learner. Otherwise why he would have put such a trivial question.@Paul $\endgroup$
    – INDIAN
    Oct 14, 2018 at 16:08
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    $\begingroup$ My problem with this proof is that $f$ is a function $X \to Y$ (and so is $g$). Thus, since only the codomain can be guaranteed to be Hausdorff, the disjoint sets $U_1$ and $U_2$ you refer to have to be in $Y$ and not in $X$. However, then you choose $y$ and z in $U_1$ and $U_2$, respectively, but this means that $y$ and $z$ are elements of $Y,$ not $X$, so the statement $f(y)>g(z)$ is meaningless since $y$ and $z$ are not necessarily in the domains of either function. $\endgroup$ Jun 8, 2019 at 1:20
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HINT: Let $h:X\to Y:x\mapsto\max\{f(x),g(x)\}$.

  1. Show that $h$ is continuous.
  2. Note that $A=\{x\in X:h(x)=g(x)\}$.
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  • $\begingroup$ This is part (b) of the problem almost word for word, my problem is part (a), I'm not sure what to make of that.. Except that they use min instead of max. $\endgroup$
    – Thoth
    Jul 1, 2012 at 20:42
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    $\begingroup$ @Nollie Tré: This is the simplest way I know to prove that $A$ is closed. In connection with (2) of the hint, you may want to recall (or prove) that if $f,g:X\to Y$ are continuous, then $\{x\in X:f(x)\ne g(x)\}$ is open in $X$; this is true for any $X$ and any Hausdorff $Y$. $\endgroup$ Jul 1, 2012 at 20:53
  • $\begingroup$ I certainly believe you, I just wonder if maybe there is an error in the book and parts (a) and (b) were switched. Anyways thanks for the help, I should hopefully be able to figure it out from here. $\endgroup$
    – Thoth
    Jul 1, 2012 at 20:57
  • $\begingroup$ @Nollie Tré: That would be my guess (that they’ve been interchanged). And the same basic argument could be used with either $\max$ or $\min$. $\endgroup$ Jul 1, 2012 at 20:58
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    $\begingroup$ @Nollie Tré: No pasting lemma is required. For any $y\in Y$, $$\begin{align*}&h^{-1}[(\leftarrow,y)]=\{x\in X:h(x)<y\}\\&=\{x\in X:f(x)<y\text{ and }g(x)<y\}=f^{-1}[(\leftarrow,y)]\cap g^{-1}[(\leftarrow,y)]\end{align*}$$ and $$\begin{align*}&h^{-1}[(y,\to)]=\{x\in X:f(x)>y\text{ or }g(x)>y\}\\&=f^{-1}[(y,\to)]\cup g^{-1}[(y,\to)]\end{align*}$$ are open, so $h$ is continuous. $\endgroup$ Jul 2, 2012 at 17:43
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Hint for the problem:

  1. Recall that $Y$ under the order topology is Hausdorff (Exercise).

  2. Show that the complement of $A$ is open. Do this by supposing that $x \notin A$. Then $g(x) < f(x)$. If this is the case, then either there exists $y$ such that $g(x) < y < f(x)$, or (exclusively) there does not exist any $y \in Y$ in between $f(x)$ and $g(x)$.

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I feel like this question could be quite difficult for a lot of beginners, so, I will post a perhaps more detailed answer than the ones I see here.

If you only wanted hints for this problem, look no further

Otherwise, here we go:

Proof: If we let $M = \{x \in X: f(x) \leq g(x)\}$, then it suffices to show that $X - M$ is open. If $X-M$ is empty, then we are done, so suppose not. Let $x \in X-M$. Then, it must be the case that $f(x) > g(x)$. We will now show the existence of a neighborhoods $U_x$ and $V_x$ about $f(x)$ and $g(x)$, respectively, that are not only disjoint but also satisfy the property that every element in $U_x$ is larger than every element of $V_x$.

Case 1: $Y$ contains no smallest or largest element

In this case, we note that there must be $a < g(x)$ and $b> f(x)$. If there is no $y \in Y$ satisfying $g(x) < y < f(x)$, then $V_x = (a, f(x))$ and $U_x = (g(x), b)$ are open sets in $Y$ satisfying the properties outlined above. If there is such $y \in Y$, then we can simply let $U_x = (y, b)$ and $V_x = (a, y)$.

Case 2: $Y$ contains a smallest element but not a largest element

We will note that this case will also cover, via a similar argument, the case where $Y$ contains a largest but not a smallest element. If $g(x)$ is not the smallest element, then there is a smallest element $a_0$ in $Y$ with $a_0 < g(x)$. Since $Y$ has no largest element, then there is $b > f(x)$. If there is no $y \in Y$ satisfying $g(x) < y < f(x)$, then we simply let $V_x = [a_0, f(x))$ and $U_x = (g(x), b)$. If there is such $y$, then let $U_x = (y,b)$ and $V_x = [a, y)$. Now, if $g(x)$ is the smallest element of $Y$, then if there is no $y$ with $g(x) < y < f(x)$, let $V_x = [g(x), f(x))$ and let $U_x = (g(x), b)$. If there is such $y$, then we simply let $V_x = [g(x), y)$ and $U_x = (y, b)$.

Case 3: $Y$ contains a largest and smallest element

If neither $g(x)$ nor $f(x)$ are the largest or smallest element, then one can proceed, as in case 1, to find $U_x$ and $V_x$. If either $g(x)$ or $f(x)$, but not both is a smallest/largest element, then one can proceed as in case 2. Thus, we consider the case where $g(x)$ is the smallest element and $f(x)$ is the largest element. If there is $y$ satisfying $g(x) < y < f(x)$, then we simply let $U_x = (y, f(x)]$ and we can let $V_x = [g(x), y)$. If no such $y$ exists, then let $U_x = (f(x), g(x)]$ and let $V_x = [f(x), g(x))$.

Since we have shown that neighborhoods $U_x$ and $V_x$ of $f(x)$ and $g(x)$, respectively, exists such that every element of $U_x$ is larger than every element of $V_x$, we see by the continuity of $f$ and $g$ that $f^{-1}(U_x)$ and $g^{-1}(V_x)$ are open in $X$. Thus, $f^{-1}(U_x) \cap g^{-1}(V_x)$ is open in $X$. If $m \in f^{-1}(U_x) \cap g^{-1}(V_x)$, then $f(m) \in U_x$ and $g(m) \in V_x$, so $f(m) > g(m)$, so $m \in X-M \implies f^{-1}(U_x) \cap g^{-1}(V_x) \subseteq X-M$. Since $x \in X-M$ was arbitrary, this shows us that for every element $x$ of $X-M$, there is a neighborhood $B_x$ of $x$ contained in $X-M$. Thus, $$X-M = \bigcup_{x \in X-M} B_x$$ So, $X-M$ is open, as desired.

As a side note, perhaps individuals with more experience will see that in the middle of the proof, we also proved that every ordered set $Y$ equipped with the order topology is Hausdorff.

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  • $\begingroup$ how does showing the existence of a neighborhoods $ U_x $and $ V_x $ about f(x) and g(x), respectively, that are not only disjoint but also satisfy the property that every element in $ U_x$ is larger than every element of $ V_x $ implies that X-M is open. Can you please explain that? $\endgroup$
    – Avenger
    Jun 23, 2020 at 4:21
  • $\begingroup$ it may seem easy to you but I am not getting it!! I will be really thankful if you explain that. $\endgroup$
    – Avenger
    Jun 23, 2020 at 4:23
  • $\begingroup$ @user795826 This is explained in full detail in the last big paragraph beginning with "Since we have shown that neighborhoods $U_x$ and $V_x$..." $\endgroup$ Jun 24, 2020 at 7:03
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Case $1$: If$ f = g $ on $X$ then $A = X $is closed

Case$ 2$ : Suppose $f\neq g$. The complement of A in X is $X − A $= {$x | g(x) < f(x)$} We will show that X − A is open. Suppose that$ X − A$ is non-empty and pick an arbitrary element $x_{0} ∈ X − A$. Let a, b and c be elements in Y such that $a ≤ g(x_0) < b ≤ f(x_0) ≤ c$

In the order topology, $[a, b)$ and $[b, c]$ are open sets. Because$ f $and $g$ are continuous , $g^{-1}([a, b))$ and $f^{-1}[b,c)$ are open in X. Their intersection is an open neighborhood of $x_0$ which is entirely contained in$ X − A$. Since $x_0 $is an arbitrary element of $X − A$, we conclude that $X − A$ is open

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  • $\begingroup$ In the order topology, $[a,b)$ is only guaranteed to be open if $a$ is the smallest element of $Y$. $(b,c]$ is only guaranteed to be open if $c$ is the largest element of $Y$. If $Y$ does now have a smallest or largest element, then neither will be open. For instance, if we consider $\mathbb{R}$ in the order topology, no half-open interval is open. $\endgroup$ Jun 8, 2019 at 0:09

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