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Let $X, Y$ be topological spaces, $Y$ an ordered set in the order topology and $f_i:X\to Y, i=1,2$ be continuous. Show that $A=\{x:f_1(x)<f_2(x)\}$ is open.

(This differs from the related questions I found in that here $Y$ has only a topological structure; in the other questions (1, 2) I found $Y$ was the real line or a metric space)

I think I was able to come up with a proof (see below), but it feels like it is more complicated than it had to be. Any ideas for a shorter, tidier alternative?


We show that $A$ is open by showing that for every $x\in A$ there exists open $U$ with $x\in U\subseteq A$.

So let $x\in A$; then $f_1(x)\not=f_2(x)$, and since $Y$ is Hausdorff we may find disjoint open $V_i\ni f_i(x)$; and then since open intervals form a basis, we may find open intervals $I_i=(a_i,b_i)\ni f_i(x)$ such that $I_i\subseteq V_i$ and thus $I_1\cap I_2=\emptyset$.

Lemma. For all $t_i \in I_i$, we have $t_1 < t_2$.

Using this lemma then, put

$$ U = \bigcap_i f_i^{-1}(I_i) $$

Since $f_i$ is continuous and $I_i$ is open, $f_i^{-1}(I_i)$ is open and then so is $U$; also since $f_i(x)\in I_i$, we have that $x\in U$, and finally, to show that $U\subseteq A$, let $u \in U$; then $f_i(u)\in I_i$ and thus by the lemma, $f_1(u) < f_2(u)$ so that $u\in A$, q.e.d.

Proof of Lemma. Let $t_i \in I_i$, and assume by way of contradiction $t_2 \le t_1$. If $t_2=t_1$ then $I_1\cap I_2\not=\emptyset$, contradiction. So try $t_2 < t_1$. In that case we have two options.

$t_1 < b_2$. In that case, from $t_2 < t_1$ and $a_2 < t_2$ (since $t_2 \in I_2$) we deduce that $t_1 \in I_2$ again contradicting $I_1\cap I_2=\emptyset$.

So $b_2 \le t_1$. Then since $f_2(x)\in I_2$ we deduce that $f_2(x) < t_1$, and since $t_1\in I_1$, we have that $t_1 < b_1$ whence we deduce that $f_2(x) < b_1$; from $f_1(x)\in I_1$ we deduce that $a_1 < f_1(x)$ and then from $f_1(x) < f_2(x)$ we deduce that $a_1 < f_2(x)$, which now means that $f_2(x) \in I_1$, again contradicting $I_1 \cap I_2 = \emptyset$.

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Here’s an approach that isn’re really essentially different but may be a little neater. Define

$$f:X\to Y\times Y:x\mapsto\langle f_1(x),f_2(x)\rangle\;.$$

If $V$ and $W$ are open sets in $Y$, $f^{-1}[V\times W]=f_1^{-1}[V]\cap f_2^{-1}[W]$ is open in $X$, so $f$ is continuous. Now let $U=\{\langle y_1,y_2\rangle\in Y\times Y:y_1<y_2\}$, the subset of $Y\times Y$ lying strictly above the diagonal; $U$ is open in $Y\times Y$, so $A=f^{-1}[U]$ is open in $X$.

To prove that $U$ is open, let $\langle y_1,y_2\rangle\in U$. If there is a $z\in(y_1,y_2)$, let $V_1=(\leftarrow,z)$ and $V_2=(z,\to)$; otherwise let $V_1=(\leftarrow,y_2)$ and $V_2=(y_1,\to)$. In either case $V_1$ and $V_2$ are disjoint open nbhds of $y_1$ and $y_2$, respectively, so $V_1\times V_2$ is an open nbhd of $\langle y_1,y_2\rangle$. Clearly $v_1<v_2$ whenever $v_1\in V_1$ and $v_2\in V_2$, so $V_1\times V_2\subseteq U$, and $U$ is open.

You could use the same idea to simplify slightly the proof of your lemma.

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  • $\begingroup$ That is indeed much neater, thanks. BTW $(z,\to)=\{z':z'>z\}$ I presume, right? $\endgroup$ May 6 '20 at 15:28
  • $\begingroup$ @whatadisgrace: You’re welcome. Yes, you understood the notation correctly; I prefer the arrow notation for rays to the more common $\pm\infty$ notation. $\endgroup$ May 6 '20 at 15:31
  • $\begingroup$ @BrianM.Scott you mean $f^{-1}[V\times W]=f^{-1}_1(V)\cap f^{-1}_2(W)$, right? $\endgroup$
    – PtF
    May 6 '20 at 15:34
  • $\begingroup$ @PtF: I sure do; thanks! $\endgroup$ May 6 '20 at 15:35

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