- #1
Tetrinity
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Homework Statement
As the title states, the problem asks to prove that the closure of the set of rational numbers is equal to the set of real numbers. The problem includes the standard definition of the rationals as {p/q | q ≠ 0, p,q ∈ Z} and also states that the closure of a set X ⊂ R is equal to the set of all its limit points.
Our lecturer suggested the following method:
- Show that the closure of Q is a subset of R.
- Show that R is a subset of the closure of Q.
- Therefore the closure of Q is equal to R
Homework Equations
I'm not sure if there's anything particularly useful here. My definition of R states that Q is a subset of R, it is ordered, the standard arithmetic operations work and it is complete (i.e. every bounded set S ⊂ R has a suprenum sup(S) ∈ R and infinum inf(S) ∈ R). It also states that every real number can be defined as a convergent sequence of rationals.
The Attempt at a Solution
I'm going to use the notation cl(X) to represent the closure of the set X.
The first step is pretty easy:
Q ⊂ R
⇒ cl(Q) ⊂ cl(R)
⇒ cl(Q) ⊂ R
(the closure of N, Z and R are already known to us at this point)
My problem lies with the second step. I know that the closure of a set X is the set of all limit points of X, and that a is a limit point to X if a is an adherent point to X\{a}. I'm just not really sure how to begin to prove the second step, and I've racked my brains to the point where I'm probably missing something obvious.
Any help with this, be it using this method or a different one, would be appreciated. Thanks.