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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 $a_x<y<b_x$. 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 $x<y$. We have $x<y<b_x$. 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 $a_x<y<b_x$. 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 $a_x<a<y<x$ 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 $x<y$.

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

Thank you.

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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, $(m,n)\subset U$.

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

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