Prove: $\sup (S \cup P) = \max \{\sup(S), \sup(T)\}$, where $S$ and $T$ are non-empty subests of $R$

Question

Prove: $\sup (S \cup P) = \max \{\sup(S),\sup(T)\}$, where $S$ and $T$ are non-empty subsets of $R$. I attempted this proof in the following manner, which I am not sure checks out:

Let $m = \sup(S \cup P)$. Then, for all $s \in S$ and $t \in T$

$$m \ge t \quad \text{and}\quad m \ge s.$$

Now, assume $\max \{\sup(S),\sup(T)\}$ is

1. $\sup(S)$

2. $\sup(T)$

If it is (1), and I let $s'=\sup(S)$, this implies that for all $s \in S$ and $t \in T$ that

$$s' \ge t \quad \text{and} \quad s' \ge s.$$

If it is (2), and I let $t'=\sup(T)$, this implies that for all $s \in S$ and $t \in T$

$$t' \ge t \quad \text{and} \quad t' \ge s.$$

I now notice that in either case, the definition of $m$ matches the definitions of $s$ or $t$, and I assume that my proof is complete. Any corrections and/or reassurance would be very much appreciated, thanks for your time.

Answer 1

Let $m_S:=\sup S$, $m_T:=\sup T$ and $m:=\sup S\cup T$.

Part1

• $S\subseteq S\cup T\implies m_S\leq m$.

• $T\subseteq S\cup T\implies m_T\leq m$.

This can captured in: $\max(m_S,m_T)\leq m$.

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Part2:

If $x\in S\cup T$ then we have the possibilities:

• $x\in S$ leading to $x\leq m_S\leq \max(m_S,m_T)$
• $x\in T$ leading to $x\leq m_T\leq \max(m_S,m_T)$.

So $\max(m_S,m_T)$ is an upper bound for $S\cup T$.

Conclusion: $m\leq\max(m_S,m_T)$.

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You proved the second part but the first part stayed under-exposed.