Post History
#4: Post edited
by
Snoopy
·
2024-01-09T01:00:26Z (11 months ago)
fixed typos: [I couldn't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly. Now it works with `A\cdot B = \\{ ab\mid a\in A, b\in B\\}`]
- > **Problem.** Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that
- $$ (\sup A)(\sup B) = \sup A\cdot B$$
- where the set of the right-hand side is defined as
- $$
A\cdot B = \{ ab\mid a\in A, b\in B\}- $$
[I can't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly.]- The problem above is a typical exercise in real analysis manipulating the definition of [supremum](https://en.wikipedia.org/wiki/Infimum_and_supremum). This is an excellent example of what Gowers called a “[fake difficulty](https://gowers.wordpress.com/2011/09/25/basic-logic-connectives-and-and-or/)” in his blog post for the undergraduate real analysis course at Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.
- > **Problem.** Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that
- $$ (\sup A)(\sup B) = \sup A\cdot B$$
- where the set of the right-hand side is defined as
- $$
- A\cdot B = \\{ ab\mid a\in A, b\in B\\}
- $$
- The problem above is a typical exercise in real analysis manipulating the definition of [supremum](https://en.wikipedia.org/wiki/Infimum_and_supremum). This is an excellent example of what Gowers called a “[fake difficulty](https://gowers.wordpress.com/2011/09/25/basic-logic-connectives-and-and-or/)” in his blog post for the undergraduate real analysis course at Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.
#3: Post edited
by
Snoopy
·
2024-01-07T22:06:06Z (11 months ago)
- > **Problem.** Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that
- $$ (\sup A)(\sup B) = \sup A\cdot B$$
- where the set of the right-hand side is defined as
- $$
- A\cdot B = \{ ab\mid a\in A, b\in B\}
- $$
- [I can't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly.]
The problem above is a typical exercise in real analysis manipulating the definition of [supremum](https://en.wikipedia.org/wiki/Infimum_and_supremum). This is an excellent example of what Gowers called a “[fake difficulty](https://gowers.wordpress.com/2011/09/25/basic-logic-connectives-and-and-or/)” in his blog post for the undergraduate real analysis course in Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.
- > **Problem.** Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that
- $$ (\sup A)(\sup B) = \sup A\cdot B$$
- where the set of the right-hand side is defined as
- $$
- A\cdot B = \{ ab\mid a\in A, b\in B\}
- $$
- [I can't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly.]
- The problem above is a typical exercise in real analysis manipulating the definition of [supremum](https://en.wikipedia.org/wiki/Infimum_and_supremum). This is an excellent example of what Gowers called a “[fake difficulty](https://gowers.wordpress.com/2011/09/25/basic-logic-connectives-and-and-or/)” in his blog post for the undergraduate real analysis course at Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.
#2: Post edited
by
Snoopy
·
2024-01-07T22:05:23Z (11 months ago)
$\sup(A\cdot B) = (\sup(A))(\sup (B))$ where $A$ and $B$ bounded sets of positive real numbers
- $\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
#1: Initial revision
by
Snoopy
·
2024-01-07T20:48:55Z (11 months ago)
$\sup(A\cdot B) = (\sup(A))(\sup (B))$ where $A$ and $B$ bounded sets of positive real numbers
> **Problem.** Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that
$$ (\sup A)(\sup B) = \sup A\cdot B$$
where the set of the right-hand side is defined as
$$
A\cdot B = \{ ab\mid a\in A, b\in B\}
$$
[I can't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly.]
The problem above is a typical exercise in real analysis manipulating the definition of [supremum](https://en.wikipedia.org/wiki/Infimum_and_supremum). This is an excellent example of what Gowers called a “[fake difficulty](https://gowers.wordpress.com/2011/09/25/basic-logic-connectives-and-and-or/)” in his blog post for the undergraduate real analysis course in Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.