# Proving $\sup{A}=\inf{U_A}$, where $U_A$ denotes set of all upper-bounds of $A$

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.

• 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 $inf U_A$ , $inf U_A \leq sup A$ ? Oct 3 '19 at 15:28
• @DeapSoup, Oh my goodness. I couldn't see this simple fact. Thanks a lot. You can answer in answer section Oct 3 '19 at 15:30
• Happy to help :) Oct 3 '19 at 15:30
• But by definition, $\sup A=\min (U_A)$. Isn't the minimum , when it exists, also the infimum? Oct 3 '19 at 15:57

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