I have hard time understanding Such That

[BarringtonT]

I have hard time understanding "Such That"

{x$\in$$Z$|x=2m+1, m$\in$$Z$}

Is this saying that x is apart of all integers and because of that x=2m+1 where m is all so apart of all integers.

if not

Is the first part x$\in$$Z$ because of the second part x=2m+1, m$\in$$Z$

or

Is second part x=2m+1, m$\in$$Z$ because of the first x$\in$$Z$

 

[yossell]

The formula asking you to take the set of all integers x such that x is identical to 2m + 1, where M itself is an integer. More simply, x is a member of the set if x an odd integer.

I'm not sure I understand your questions. By 'apart' did you mean 'a part'? If so, it might be better to say "x is an intger". More importantly, there's no implication that x should be odd because x is an integer. SUCH THAT is not a causal connective.

Indeed, one could have specified the same set without the opening clause that x is an integer, since this in fact follows from the fact that x is an odd integer.

 

[Mark44]

"Such that" is used before one or more qualifiers that limit what we're discussing.

Here are some examples.

1. {x $\in$ Z} - this set represents all of the integers, negative, zero, and positive.
2. {x $\in$ Z | x > 2} - The bar (|) can be read as "such that." The inequality limits the set so that we're now talking about only the integers larger than 2.

3. {x $\in$ Z | x > 2 and x < 7} - We're further limiting the set, so now we're considering only {3, 4, 5, 6}. This could also be written as {x $\in$ Z | 2 < x < 7}, and means exactly the same thing.
4. {x $\in$ Z | x > 2 and x < 3} - We have limited things so much that there are no members in the set, because there are no integers that are both larger than 2 and smaller than 3.