# Seemingly simple problem

1. May 8, 2008

### Diffy

Let $$R$$ be the set of real numbers and let $$<, \leq$$ have their usual meanings.

Show that $$\{ x \in R : 0 \leq x \leq 3\} \cap \{ x \in R : -1 < x < 1\} = \{ x \in R : 0 \leq x < 1\}$$

This seems very straight forward, but I am having trouble with wording and figuring out what is sufficient enough to conclude the proof.

So here is what I would argue.
By the definition of the intersection of two sets we have
$$\{ x \in R : 0 \leq x \leq 3\} \cap \{ x \in R : -1 < x < 1\} = \{ x \in R : 0 \leq x \leq 3 and -1 < x < 1 \}$$

Here is where I am not sure about the language that I need to use...

Can I just say something like:

The only way both conditions can be true is if $$0 \leq x < 1$$

Therefore we have $$\{ x \in R : 0 \leq x \leq 3 and -1 < x < 1 \} = \{ x \in R : 0 \leq x < 1\}$$

Or should I not treat this as obvious???

Or is there a better way to approach this all together?

My definition of intersection is as follows:

$$X \cap Y = \{x : x\in X and x\in Y\}$$

Up to this point there is no formal treatment of either open and closed intervals, so I am not sure how informal or obvious I should assume them to be.

2. May 8, 2008

### Bob3141592

I don't know how if this notion fits your formality target, but I think the idea of open and closed intervals is already embedded in the $$<$$ and $$\leq$$ operators. When a mathematical proof uses a statement like "have their usual meaning", I'd guess a little informality would be tolerated.

3. May 9, 2008

### tiny-tim

Hi Diffy!

I suspect it's a logic test, and they want you to say something like …

if x > a and x > b, then x > sup{a,b}.

4. May 9, 2008

### BryanP

I guess you can break it up further to show why exactly it's like that.

{x in R | 0 <= x <= 3} int {x in R | -1 < x < 1 }

= [{x in R | x >= 0) int {x in R | x <= 3}] int [{x in R | x > -1} int {x in R | x < 1}] <-- Break it down to separate intersections of sets

= [{x in R | x >= 0) int {x in R | x > -1}] int [{x in R | x <= 3} int {x in R | x < 1}] <-- Rearrange the sets since we're dealing with all set intersections

= {x in R | x >= 0) int {x in R | x < 1} <--simplify to satisfy set conditions when intersected

= {x in R | 0 <= x < 1} <-- simplify further to satisfy set conditions when intersected

5. May 9, 2008

### Diffy

Hi everyone,

Tiny Tim, that seems to make a lot of sense. Even though the author never uses Sup or Inf, I will go ahead an proceed that way.

I am still a bit confused though on having "less than" and "Less than or equal to"

So I will say since x < 1 and $$x \leq 3$$, then we must have x ? inf(1, 3)

where the question mark is < or $$\leq$$?

BryanP, I think you are saying the same thing that TinyTim is, so thanks for the help, I think breaking it up into that many pieces is missing the main point of determining how we know that the intersections of these intervals are what they are.

BobPi,
I would have rounded 3141592 up to 3141593.

Anyways, you are right, I think I can be very informal, but I was just missing the piece on how we know if x<1 and x < 3 that x must be less than 1, using least upper bounds and greatest lower bounds is the key I think.

Last edited: May 9, 2008
6. May 9, 2008

### tiny-tim

Hi Diffy!
For the "≤"? Just copy-and-paste what someone else has done!
ah … you need something like lowest-common-multiple in fractions.

Try rewriting one of the sets {x < 1} and {x ≤ 3} so that it matches the other …

{x < 1} = {x ≤ 1} ∩ {…?}
But Bob is modest, and a model of understatement!

7. May 9, 2008

### Diffy

Nah, I forgot a / in my tex tag, so the whole post looked very strange.

Alrighty,
{x < 1} = {x ≤ 1} ∩ {x < 1}?

or

{x < 1} = {x ≤ 1} ∩ {x ≤ ?}

I don't see where you are going.. Sorry I'm a little slow and I have trouble these details.

lol I know, I was just teasing, all in fun.

8. May 9, 2008

### tiny-tim

Hi Diffy!

{x < 1} = {x ≤ 1} ∩ {x ≠ 1} ?