MHB Expressing a set as the difference between two sets.

[rayne1]

Let a, b, c, and d be real numbers with a < b < c < d. Express the set [a, b]U[c, d] as the difference of two sets.

I know that [a,b]U[c,d] is a union and what a difference of two sets is, but I don't quite understand this question.

 

[mathbalarka]

Note that you're given the condition $a < b < c < d$. Thus, $[a, b]$ and $[c, d]$ are disjoint, i.e., $(b, c)$ is non-null. Consider the compactfied interval $[a, d] \supset [a, b] \cup [c, d]$. Can you "subtract" something from $[a, d]$ to produce $[a, b] \cup [c, d]$?

 

[evinda]

Let a, b, c, and d be real numbers with a < b < c < d. Express the set [a, b]U[c, d] as the difference of two sets.

I know that [a,b]U[c,d] is a union and what a difference of two sets is, but I don't quite understand this question.

Hi!

$$[a, b] \cup [c, d]=[a,d] \setminus (b,c)$$

since at the interval $[a, b] \cup [c, d]$ are included all the numbers from the interval $[a,d]$, except the ones that belong to the open interval $(b,c)$.

 

[rayne1]

Note that you're given the condition $a < b < c < d$. Thus, $[a, b]$ and $[c, d]$ are disjoint, i.e., $(b, c)$ is non-null. Consider the compactfied interval $[a, d] \supset [a, b] \cup [c, d]$. Can you "subtract" something from $[a, d]$ to produce $[a, b] \cup [c, d]$?

So like what evinda said, (b,c)?

 

[mathbalarka]

Yes, well, evinda revealed it. It can be written as $[a, d] - (b, c)$.

 

[Deveno]

In general, a union of bounded intervals can be expressed as the interval bounded by the inf and the sup of all the intervals, minus the "holes".

I like to think of $A - B$ as: $A$, except for the bite $B$ took out of $A$.

 

[Evgeny.Makarov]

Let a, b, c, and d be real numbers with a < b < c < d. Express the set [a, b]U[c, d] as the difference of two sets.

And, of course, $[a, b]\cup[c, d]=([a, b]\cup[c, d])\setminus\emptyset$. Many other variants are possible. If the problem asked to represent $[a, b]\cup[c, d]$ as the difference of two segments, then I believe there is only one solution.