# prove for every open bounded subset of R, the largest open interval exists

Show that if G is an open bounded subset of real numbers, and if x belongs to G, then there exists the largest open interval $$I_x$$ containing x such that $$I_x$$ is a subset of G.

I know that every open subset can be written as countable union of mutually disjoint open intervals. so since G is open and G is the largest subset of itself and G contains x, then the largest open interval containing x, exists which is G itself.
I know my statement is not a real proof. the main concept of the question to me is confusing. I would appreciate any idea about the proof for this question.

Let $$U_1=\{y\in G:[x,y)\subseteq G\}.\ U_1$$ is non-empty because $$G$$ is open. And since $$G$$ is bounded, $$z:=\sup U_1$$ exists and is finite. Now, $$z\notin U_1$$ but $$x\le y satisfies $$y\in U_1$$ (why?), so the interval $$[x,z)$$ is maximal.

Set $$U_2=\{y\in G:(y,x]\subseteq G\}$$, repeat the above argument setting $$w=\inf U_2$$.

It follows that $$U=U_1\cup U_2=(w,z)\subseteq G$$ is the maximal open interval containing $$x$$.

For $$x,y\in \Bbb R$$ let $$In[x,y]=[\min(x,y),\max (x,y)].$$ Show that $$In[x,y]\cup In[x,z]=[\min(x,y,z),\max(x,y,z)]\supset In[y,z]$$ for all $$x,y,z\in \Bbb R.$$

Any bounded $$S\subset \Bbb R$$ is an interval iff $$S$$ is convex iff $$\forall y,z\in S\,(In[y,z]\subset S).$$

$$(1).$$ Let $$I_x=\cup \{In[x,y]: y\in \Bbb R\land In[x,y]\subset G\}.$$

Obviously $$I_x\subset G$$ and hence also $$I_x$$ is bounded.

$$(2).$$ To show that $$I_x$$ is an interval:

Take any $$y,z\in I_x$$. There exist $$y',z'$$ such that $$y\in In[x,y'] \subset G$$ and $$z\in In[x,z']\subset G.$$ So $$In[x,y']$$ and $$In[x,z']$$ are subsets of $$I_x.$$

We have $$\min(x,y')\le y\le \max (x,y')$$ and $$\min (x,z') \le z \le \max (x,z'),$$ which implies $$\min(x,y',z') \le \min(y,z)\le \max(y,z) \le \max (x,y',z'),$$ so $$In[y,z]\subset [\min (x,y',z'),\max(x,y',z')]=In[x,y']\cup In[x,z']\subset I_x.$$

$$(3).$$ Let $$U=\sup I_x$$ and $$L=\inf I_x,$$ which exist because $$I_x\ne \emptyset$$ (because $$In[x,x]=\{x\}\subset G$$) and because $$I_x$$ is bounded. Since $$I_x$$ is a bounded interval, we have $$(L,U)\subset I_x\subset [L,U].$$

By contradiction, assume $$U\in G.$$ Since $$G$$ is open,there exists $$y>U$$ such that $$[U,y]\subset G.$$ And we have $$[x,U)\subset I_x\subset G,$$ so $$In[x,y]=[x,y]=[x,U)\cup [U,y]\subset G,$$ implying $$[x,y]\subset I_x .$$ But then $$\sup I_x\ge y>U=\sup I_x,$$ which is absurd. Therefore $$U\not \in G.$$ Similarly we show that $$L\not \in G.$$ Therefore $$I_x=(L,U).$$ So $$I_x$$ is open.

$$(4).$$ Finally, let $$J$$ be $$any$$ interval such that $$x\in J\subset G.$$ Then $$\sup J\le U,$$ otherwise $$U\in [x,U]\subset [x,\sup J)\subset G,$$ implying $$U\in G.$$ Similarly $$\inf J\ge L.$$ Therefore $$J=J\cap G\subset [L,U]\cap G=(L,U)=I_x.$$

• Assume that we have shown that if $C$ is any family of real intervals such that $x\in \cap C,$ then $\cup C$ is convex. Then we can define $I_x=\cup C$ where $I\in C$ iff $I$ is an open interval such that $x\in I\subset G.$ Then Step $(2)$ in my Answer is immediate, and the rest can stand verbatim. – DanielWainfleet Jan 27 at 7:06