Prove that the intersection of two open intervals in $\mathbb{R}$ is either empty or an open interval in $\mathbb{R}$

Question

That's how I tried, is this valid ?

An open interval in $\mathbb{R}$ assumes one of the following forms: $(-\infty,a)$, $(b,\infty)$, $(c,d)$ or $(-\infty,\infty)$.

Since $(-\infty,a)\subset(-\infty,\infty)$, $(b,\infty)\subset(-\infty,\infty)$ and $(c,d)\subset(-\infty,\infty)$, their intersections are going to be open intervals. Therefore, we only need to verify these intersections: $(-\infty,a)\cap(b,\infty)$, $(-\infty,a)\cap(c,d)$ and $(b,\infty)\cap(c,d)$.

Case $(-\infty,a)\cap(b,\infty)$:

Assume $b<a$.

If $x\in(-\infty,a)\cap(b,\infty)$, then $x<a$ and $x<b$. Then $b<x<a$, so $x\in(b,a)$. Since x was arbitrary, we've shown that $(b,a)\subset(-\infty,a)\cap(b,\infty)$. Now, if $x\in(b,a)$, then $b<x<a$. From this, $x<a$, then $x\in(-\infty,a)$. Also, $b<x$, then $x\in(b,\infty)$. Therefore, $x\in(-\infty,a)$ and $x\in(b,\infty)$. So, $x\in(-\infty,a)\cap(b,\infty)$. Since x was arbitrary, we've shown that $(-\infty,a)\cap(b,\infty)\subset(b,a)$. Finally, $(-\infty,a)\cap(b,\infty)=(b,a)$, with $(b,a)$ an open interval.

Assume $a<b$.

If $x\in(-\infty,a)\cap(b,\infty)$, then $x<a$ and $x<b$. Then $b<x<a$, so $b<a$. But $a<b$, therefore $(-\infty,a)\cap(b,\infty)=\emptyset$.

Case $(-\infty,a)\cap(c,d)$:

Assume $d<a$.

In this case $(c,d)\subset(-\infty,a)$ and then $(-\infty,a)\cap(c,d)=(c,d)$

Assume $c<a$ and $a<d$

If $x\in(-\infty,a)\cap(c,d)$, then $x<a$ and $c<x<d$. Then $x<a$ and $c<x$ and $x<d$. Then $c<x<a$ and $x<d$. Then $x\in(c,a)$. Since x was arbitrary, we've shown that $(c,a)\subset(-\infty,a)\cap(c,d)$. Now, if $x\in(c,a)$, then $c<x<a$. Then $x\in(-\infty,a)$. Also, since $a<d$, $c<x<a<d$, then $c<x<d$. Therefore $x\in(c,d)$. Then $x\in(-\infty,a)$ and $x\in(c,d)$. This way $x\in(-\infty,a)\cap(c,d)$. Since x was arbitrary, we've shown that $(-\infty,a)\cap(c,d)\subset(c,a)$. Finally $(-\infty,a)\cap(c,d)=(c,a)$, with $(c,a)$ an open interval.

Assume $a<c$ and $a<d$.

If $x\in(-\infty,a)\cap(c,d)$, then $x<a$ and $c<x<d$. Then $x<a$ and $c<x$ and $x<d$. Then $c<x<a$ and $x<d$. Then $c<a$. But $a<c$, therefore $(-\infty,a)\cap(c,d)=\emptyset$.

Case $(b,\infty)\cap(c,d)$:

Assume $b<c$.

In this case $(c,d)\subset(b,\infty)$ and then $(b,\infty)\cap(c,d)=(c,d)$

Assume $c<b$ and $b<d$.

If $x\in(b,\infty)\cap(c,d)$, then $b<x$ and $c<x<d$. Then $b<x$ and $c<x$ and $x<d$. Then $b<x<d$ and $c<x$. Then $x\in(b,d)$. Since x was arbitrary, we've shown that $(b,d)\subset(b,\infty)\cap(c,d)$. Now, if $x\in(b,d)$, then $b<x<d$. Then $x\in(b,\infty)$. Also, since $c<b$, $c<b<x<d$, then $c<x<d$. Therefore $x\in(c,d)$. Then $x\in(b,\infty)$ and $x\in(c,d)$. This way $x\in(b,\infty)\cap(c,d)$. Since x was arbitrary, we've shown that $(b,\infty)\cap(c,d)\subset(b,d)$. Finally $(b,\infty)\cap(c,d)=(b,d)$, with $(b,d)$ an open interval.

Assume $c<b$ and $d<b$.

If $x\in(b,\infty)\cap(c,d)$, then $b<x$ and $c<x<d$. Then $b<x$ and $c<x$ and $x<d$. Then $b<x<d$. So, $b<d$. But $d<b$, therefore $(b,\infty)\cap(c,d)=\emptyset$.

Answer 1

Let $I_1=]a_1,b_1[$ and $I_2=]a_2,b_2[$

It is enough to look at the case $I_1\cap I_2\ne \emptyset$. If $x\in I_1\cap I_2\ne \emptyset$ then there is an open interval $J_1=]x-\epsilon_1,x+\epsilon_2[$ contained in $ I_1$ and an open interval $J_2=]x-\epsilon_3,x+\epsilon_4[$ contained in $I_2$ where the four epsilons are positive. Taking $\alpha=\max\{\epsilon_1,\epsilon_3\}$ and $\beta=\min\{\epsilon_2,\epsilon_4\}$ we get the open interval $]x-\alpha,x+\beta[$ contained in $I_1\cap I_2$. We are done for bounded open intervals and similar reasoning for unbounded ones.

Answer 2

An open interval on $\mathbb{R}$ is one that does not contain all of its accumulation points, namely its two endpoints.

So lets suppose that $A = (a_0,a_1)$ and $B=(b_0,b_1)$ are our two open intervals, and only consider the case when $A\cap B\neq\emptyset$.

Let $a_0\lt a_1 $ and $b_0\lt b_1 $, and WLOG suppose $a_0\lt b_0 $, then since the intersection is non-empty, we can say, $b_0\lt a_1 $. Now, it will suffice to show that $b_0$ and $a_1$ are accumulation points of $A\cap B$, and that $b_0,a_1\notin A\cap B$. If we prove this, then the open interval $(b_0,a_1) = A\cap B$.

Consider the open neighborhood $(b_0-\varepsilon,b_0+\varepsilon)$, where $\varepsilon\in\mathbb{R}$ is arbitrary. Lets pick a value $\beta$ such that $b_0\lt\beta\lt b_1$ and $\beta\lt\varepsilon$. Then $\beta\in (b_0-\varepsilon,b_0+\varepsilon)$, and $\beta\in B$ as an immediate consequence of $b_0\lt\beta\lt \varepsilon$. Also, $\beta\in A\cap B$ as an immediate consequence of $b_0\lt\beta\lt b_1$. Thus for any neighborhood of $b_0$ we can find some $\beta$ that is in the neighborhood and in $A\cap B$, which means by definition, $b_0$ is an accumulation point of $A\cap B$. We can use similar logic to show that $a_1$ is an accumulation point of $A\cap B$.

Now we have to show that $b_0,a_1\notin A\cap B$. This follows from the fact that $A, B$ are open and do not contain their endpoints, which gives us that $b_0\notin B$ and $a_1\notin A$. Hence, we get $b_0,a_1\notin A\cap B$.

If you must prove that this open set $A\cap B$ is an interval, then it follows from the fact that all points $b\in B$ are such that $b\gt b_0$, and similarly for $A$. So all points in $A\cap B$ must be between $b_0$ and $a_1$

Thus, $A\cap B$ does not contain all of its accumulation points and is consequently an open set, namely the open interval $(b_0,a_1)$.

For the cases where the intervals $A,B$ are unbounded, you should use the method outlined by user Piquito, since we can't argue on the endpoints like this if they are infinite. But either way it is better in this case to do a more general proof than list a lot of cases exhaustively, in my opinion.