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Let $A$ be a subset of $B$, where $A$ is nonempty. How can I show that $\sup A \leq \sup B$?

My attempt

I said let element $a$ be in $A$, which means $a$ is in $B$ because $A$ is a subset of $B$. I also stated that $\sup A$ is an upper bound of $A$, so for all elements $a \in A$, $a \leq \sup A$. Also if $\sup B$ is an upper bound of $B$, $b \leq \sup B$ for all elements $b$. How can use these to arrive at the result I desire?

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  • $\begingroup$ TeX tips: use command \in and \subset (or \subseteq) to type appropriate notation. For more, see math notation guide. $\endgroup$
    – user147263
    Sep 6, 2014 at 20:42

2 Answers 2

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Since $A \subset B$, $\sup B$ is an upper bound for $A$. Since $\sup A$ is the least upper bound for $A$ by definition, it must be less than or equal $\sup B$.

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    – Newb
    Sep 6, 2014 at 20:42
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Since this is under Real Analysis, I would assume $A \subset B \subset \mathbb{R}$ and given, that $A$ is non-empty. I will give a longer proof (which is superfluous to be honest) but it will give you some confidence in seeing how proof works.

We proof by contradiction, assume $\sup A > \sup B$. Now suppose that $A, B$ both bounded above.

Then let $\alpha = \{U_1, U_2, U_3, \ldots \}$ be the set of the upper bounds on $A$.

Similarly, let $\beta = \{V_1, V_2, V_3, \ldots\}$ be the set of upper bounds on $B$.

Our first observation: $$\min \alpha \leq \min \beta$$

Now let $\alpha \cap \beta = \{W_1, W_2, W_3, \ldots\}$, then we have our second observation:

$$\min(\alpha \cap \beta) = \min(\alpha)$$

Third observation: $\alpha \cap \beta$ must contain the least upper bound (supremum) of $A$ i.e. $$\sup A \in \alpha \cap \beta$$

In particular, $$\sup A = \min(\alpha \cap \beta)$$

By second observation, we further have (call this fourth observation):

$$\sup A = \min(\alpha \cap \beta) = \min(\alpha)$$

And note that we have $\sup B = \min \beta$. Now, since $\min \alpha \leq \min \beta$ by first observation, using our fourth observation, we have that:

$$\min(\alpha \cap \beta) \leq \min \beta$$

Since by third observation, $\sup A = \min(\alpha \cap \beta)$ and $\sup B = \min \beta$. We have that:

$$\sup A \leq \sup B$$

which is a contradiction to our assumption, so it must be wrong.

Comment: I realize that defining $\alpha, \beta, \alpha \cap \beta$ as a set with showing their elements explicitly is of no use to this proof, but it is there to give a sense to some readers who need to visualize it via a diagram for example. One can imagine that $A$ is a small circle in $B$ (as $A \subset B$) and that elements of $\alpha$ i.e. $U_1, U_2, U_3$ etc. are points in the circle outside of $A$ and beyond (but could also be in $A$ itself), whereas elements of $\beta$ i.e. $V_1, V_2, V_3$ etc. are points in the circle outside of $B$ and beyond (but could also be in $B$ itself). Elements of $\alpha \cap \beta$ i.e. $W_1, W_2, W_3$ etc. are just the intersection of these two sets.

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