Suprema proof: quick check please

[tylerc1991]

Homework Statement

Assume that $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$ are each bounded above. Then the set
$Z = \{ x + y : x \in X, y \in Y \}$
is bounded above, and
$\text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y). \quad \quad$

The Attempt at a Solution

Let $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$ be bounded above, and define the set $Z$ as
$Z = \{ x + y : x \in X, y \in Y \}.$
Let $e_x$ represent an element of $X$, $e_y$ represent an element of $Y$, and $e_z$ represent an element of $Z$.
For any $e_z$, there exists some $e_x$ and $e_y$ such that
$e_z = e_x + e_y. \quad \quad (1)$
Since $X$ and $Y$ are bounded above, there exists a least upper bound for $X$ and $Y$, which can be represented as $x_{sup} = \text{sup}(X) \text{ and } y_{sup} = \text{sup}(Y)$.
By the definition of the least upper bound, for all $e_x \in X, \, x_{sup} \geq e_x$.
Similarly, for all $e_y \in Y, \, y_{sup} \geq e_y$.
From equation (1), this means that
$e_z = e_x + e_y \leq x_{sup} + y_{sup} \quad \quad (2)$
for all $e_x$ and $e_y$.
From equation (2), we see that $x_{sup} + y_{sup}$ is an upper bound for $Z$.
By definition, this proves that $Z$ is bounded above.
Since $Z$ is bounded above, note that $Z$ has a least upper bound which can be represented as $z_{sup} = \text{sup}(Z)$.
By the definition of a least upper bound, $z_{sup} \leq x_{sup} + y_{sup}$, or
$\text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y).$ Q.E.D.

It wanted to be sure that this is 100% good to go before I have to turn this in. Thank you very very much for anyone that takes a look!

 

[icystrike]

Looks good to me. Wait for more comments

 

[micromass]

Homework Statement

Assume that $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$ are each bounded above. Then the set
$Z = \{ x + y : x \in X, y \in Y \}$
is bounded above, and
$\text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y). \quad \quad$

The Attempt at a Solution

Let $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$ be bounded above, and define the set $Z$ as
$Z = \{ x + y : x \in X, y \in Y \}.$
Let $e_x$ represent an element of $X$, $e_y$ represent an element of $Y$, and $e_z$ represent an element of $Z$.
For any $e_z$, there exists some $e_x$ and $e_y$ such that
$e_z = e_x + e_y. \quad \quad (1)$
Since $X$ and $Y$ are bounded above, there exists a least upper bound for $X$ and $Y$, which can be represented as $x_{sup} = \text{sup}(X) \text{ and } y_{sup} = \text{sup}(Y)$.
By the definition of the least upper bound, for all $e_x \in X, \, x_{sup} \geq e_x$.
Similarly, for all $e_y \in Y, \, y_{sup} \geq e_y$.
From equation (1), this means that
$e_z = e_x + e_y \leq x_{sup} + y_{sup} \quad \quad (2)$
for all $e_x$ and $e_y$.
From equation (2), we see that $x_{sup} + y_{sup}$ is an upper bound for $Z$.
By definition, this proves that $Z$ is bounded above.
Since $Z$ is bounded above, note that $Z$ has a least upper bound which can be represented as $z_{sup} = \text{sup}(Z)$.
By the definition of a least upper bound, $z_{sup} \leq x_{sup} + y_{sup}$, or
$\text{sup}(Z) \leq \text{sup}(X) + \text{sup}(Y).$ Q.E.D.

It wanted to be sure that this is 100% good to go before I have to turn this in. Thank you very very much for anyone that takes a look!

Very nicely done! The proof is good, but it's also written down superbly!
This would exactly be the kind of proof that I would like to receive as an assignment!

 

[tylerc1991]

Very nicely done! The proof is good, but it's also written down superbly!
This would exactly be the kind of proof that I would like to receive as an assignment!

Thank you very much for the encouraging feedback! My only small concern was the way I brought in the fact that e_z = e_x + e_y for some e_x and e_y. Intuitively, this is pretty obvious, but the course requires near perfect style as well as great logic.

 

[micromass]

Thank you very much for the encouraging feedback! My only small concern was the way I brought in the fact that e_z = e_x + e_y for some e_x and e_y. Intuitively, this is pretty obvious, but the course requires near perfect style as well as great logic.

I wouldn't worry about that, it's pretty obvious.

Maybe you will want to explain (2) some more. It's also obvious, but your professor might complain there...