# If A is in B then sup(A) < sup(B)

## Homework Statement

Let $X$ and $Y$ be nonempty subsets of real numbers such that $X \subseteq Y$ and $Y$ is bounded above. Prove that $\sup X \le \sup Y$

## The Attempt at a Solution

Case 1: $X = Y$. Trivially, $\sup X \le \sup Y$.

Case 2: $X \subset Y$. Then there exists a $y^* \in Y$ that is an upper bound for $X$. Since $\sup X$ is the least upper bound, $\sup X \le y^*$. But $y^* \le \sup Y$ since $\sup Y$ is an upper bound for $Y$. So $\sup X \le y^* \le \sup Y$, and so $\sup X \le \sup Y$

## Answers and Replies

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fresh_42
Mentor
Why should $y^*$ (= any upper bound) be less than $\operatorname{sup}Y$ (= least upper bound)? The contrary seems right, especially since I cannot see any reason why $y^*\in Y$. However, you don't need $y^*$ at all. Why not work directly with $\operatorname{sup}Y$ as upper bound for both?