# If $A$ is a subset of $B$, then $\sup A \leq \sup B$

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?

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

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 $\sup B$.

• $\checkmark\phantom{}$
– Newb
Sep 6, 2014 at 20:42

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.