Let [tex]R[/tex] be the set of real numbers and let [tex] <, \leq [/tex] have their usual meanings.(adsbygoogle = window.adsbygoogle || []).push({});

Show that [tex] \{ x \in R : 0 \leq x \leq 3\} \cap \{ x \in R : -1 < x < 1\} = \{ x \in R : 0 \leq x < 1\} [/tex]

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

[tex] \{ 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 \} [/tex]

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 [tex]0 \leq x < 1[/tex]

Therefore we have [tex]\{ x \in R : 0 \leq x \leq 3 and -1 < x < 1 \} = \{ x \in R : 0 \leq x < 1\}[/tex]

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:

[tex] X \cap Y = \{x : x\in X and x\in Y\} [/tex]

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.

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# Seemingly simple problem

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