# Show that $\{x : f_1(x) < f_2(x)\}$ is open

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) 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 $$I_1 \cap I_2 = \emptyset$$.

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 $$U=\{\langle y_1,y_2\rangle\in Y\times Y:y_1, the subset of $$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 $$v_1 whenever $$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.

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