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Consider a non-empty set $A\subset\mathbb{R}$ that is bounded above. Denote by $U_A$, the set of all upper-bounds of $A$.

I would like to prove that $\sup{A}=\inf{~U_A}$. (I think this must hold)

$\underline{My~Approach}:$
By order-completeness, $\sup{A}$ exists in $\mathbb{R}$ and $\inf{U_A}$ exists in $\mathbb{R}$.
$\forall x\in A, x$ is lower-bound of $U_A$.
$\implies\forall x\in A, x\le\inf{U_A}$. (by definition of infimum)
$\implies\inf{U_A}$ is upper-bound of $A$.
$\implies\sup{A}\le\inf{U_A}$. (by definition of supremum)

Also, $\forall y\in U_A, y$ is upper-bound of $A$.
$\implies\forall y\in U_A,\sup{A}\le y$. (by definition of supremum)
$\implies\sup{A}$ is lower-bound of $U_A$.
$\implies\sup{A}\le\inf{U_A}$. (by definition of infimum)

From both arguments, I get the same inequality, i.e., $\sup{A}\le\inf{U_A}$.
If I somehow establish that $\inf{U_A}\le\sup{A}$, then I am done. But, I don't know how to establish this.
I am stuck here.

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    $\begingroup$ A mere freshman undergrad math student here, but doesn't $ inf U_A \leq sup A $ hold, since $ sup A \in U_A $, so by definition of $ inf U_A $ , $ inf U_A \leq sup A $ ? $\endgroup$
    – DeapSoup
    Oct 3 '19 at 15:28
  • $\begingroup$ @DeapSoup, Oh my goodness. I couldn't see this simple fact. Thanks a lot. You can answer in answer section $\endgroup$
    – spkakkar
    Oct 3 '19 at 15:30
  • $\begingroup$ Happy to help :) $\endgroup$
    – DeapSoup
    Oct 3 '19 at 15:30
  • $\begingroup$ But by definition, $\sup A=\min (U_A)$. Isn't the minimum , when it exists, also the infimum? $\endgroup$
    – Bernard
    Oct 3 '19 at 15:57
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You get inf $ U_A \leq $ sup A, since sup A $ \in U_A $, so by definition of inf $ U_A $, inf $ U_A \leq $ sup A

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