# $U\subset\mathbb R$ is open, then $\forall x\in U(\inf\{a:(a,x]\subset U\},\sup\{b:[x,b)\subset U\}) \subset U$

If $$U\subset\mathbb R$$ is open, $$x\in U$$, then $$I_x = (\inf\{a:(a,x]\subset U\},\sup\{b:[x,b)\subset U\}) \in U$$

My work (1):

Let $$a_x = \inf\{a:(a,x]\subset U\}$$ and $$b_x = \sup\{b:[x,b)\subset U\}$$. Let $$y\in I_x$$, so $$a_x. If $$y=x$$, we are done, so let $$y\neq x$$. Assume $$y < x$$. In this case, $$x > y > a_x$$ proves useful. Now if $$a_x \in \{a:(a,x]\subset U\}$$ then $$(a_x,x] \subset U$$ and so $$y\in U$$. What happens if $$a_x \notin \{a:(a,x]\subset U\}$$?

Similarly, assume $$x. We have $$x. If $$b_x\in\{b:[x,b)\subset U\}$$ then $$[x,b_x)\subset U$$ and $$y\in U$$. What happens if $$b_x\notin\{b:[x,b)\subset U\}$$?

My work (2):

Let $$a_x = \inf\{a:(a,x]\subset U\}$$ and $$b_x = \sup\{b:[x,b)\subset U\}$$. Let $$y\in I_x$$, so $$a_x. If $$y=x$$, we are done, so let $$y\neq x$$. Assume $$y < x$$. In this case, $$x > y > a_x$$ proves useful. There exists some $$a$$ such that $$a_x such that $$a\in \{a:(a,x]\subset U\}$$. This is because if such an $$a$$ doesn't exist, then $$y$$ will be a lower bound for $$\{a:(a,x]\subset U\}$$, which contradicts the fact that $$a_x$$ is its infimum. Now $$(a,x]\subset U \implies y\in U$$.

Similarly for the case $$x.

While (2) seems fine, I am unable to complete my thoughts in (1).

Thank you.

## 1 Answer

Pick your favorite $$x \in U$$. Since we know that open sets in $$\mathbb{R}$$ are a countable union of disjoin open intervals, this $$x$$ must fall in some interval $$(m,n)$$. Remains to show that $$inf\{...\} = m$$. Let $$\epsilon>0$$. Suppose $$inf\{...\} = m - \epsilon$$, then contradiction because that means $$m\in U$$. Suppose $$inf\{...\} = m + \epsilon$$, then contradiction because there exists an interval starting from $$m+\epsilon/2$$ that is the candidate for being the inf. Therefore, $$inf\{...\} = m$$ and a symmetric argument shows $$sup\{...\} = n$$. And clearly, $$(m,n)\subset U$$.

• I think you have misunderstood. What is your choice of $x\in U$ used for constructing $I_x$? Your argument seems incorrect. – epsilon-emperor Feb 2 at 4:31
• ah, i misread. my bad – Scott Hahn Feb 2 at 4:32
• updated my answer – Scott Hahn Feb 2 at 4:43