# For $X,Y \subseteq \Bbb R$ define $X+Y : \{x +y \mid x \in X, y \in Y\}.$ An example where $X +Y \neq \Bbb R$.

For $$X,Y \subseteq \Bbb R$$ define $$X+Y : \{x + y \mid x \in X, y \in Y\}$$ examples where $$X + Y \neq \Bbb R$$ are

(A.) $$X = \Bbb Q$$, $$Y = \Bbb R \setminus \Bbb Q$$

(B.) $$X = \Bbb Z$$, $$Y = [-1/2,1/2]$$

(C.) $$X = (-\infty,100]$$, $$Y = \{p \in N \mid p \text{ is prime} \}$$

(D.) $$X=(-\infty, 100]$$, $$Y = \Bbb Z$$

I think option $$A$$ is correct, since we can get irrationals by summing $$X$$ and $$Y$$, but I couldn't find a way to get rationals.

Option $$B$$ is incorrect since sum will be of the form $$.... [-3/2,-1/2], [-1/2,1/2], [1/2,3/2]....$$ which covers the whole set $$\Bbb R$$.

Option $$D$$ is also incorrect.

Option $$C$$ seems incorrect by intuition, but I am not sure about $$C$$. I'm not able to get clear idea about the type of sets option $$C$$ would form. Any suggestion for $$C ?$$

• btw use \mathbb Z \mathbbQ \mathbb R otherwise they look like regular sets. – qwr Aug 14 at 4:48
• Option C is interesting because most primes are not separated by gaps of 100 but infinitely large gaps do exist – qwr Aug 14 at 4:49
• It’s “$ℝ \setminus ℚ$”, not “$ℝ / ℚ$”. The corresponding tex command even is \setminus. – k.stm Aug 14 at 5:52

In (A) we cannot write a rational number $$x$$ as $$y+z$$ with $$x$$ rational and $$y$$ irrational because $$z=x-y$$ would then be rational. In fact $$X+Y$$ consist only of irrational numbers.

(C) and (D) are both incorrect. Given any real number $$x$$ there is a 'large' prime number $$p$$ such that $$x-y \leq 100$$. Hence $$x=(x-p)+p \in (-\infty, 100] + P$$ where $$P$$ is the set of all prime numbers.

You are correct about 1). Any irrational + rational will always be an irrational so $$\mathbb Q + \mathbb {IR}\subset \mathbb {IR} \subsetneq \mathbb R$$.

(I'm using the notation $$\mathbb {IR}:= \mathbb R\setminus \mathbb Q$$ only because I think it is easy to read in this instance. It might not be a universal symbol.)

And you are correct about 2). The intervals $$[n-\frac 12, n+\frac 12]$$ "cover" the reals so that for any real $$z$$ there is some $$n\in \mathbb Z$$ (possibly two) so that $$n-\frac 12\le z \le n+\frac 12$$. If we let $$x = n$$ and $$y=z -n$$ we get $$x \in X$$ and $$y\in Y$$ and $$z = X+Y$$ so $$\mathbb R\subset X+Y$$. (And as $$\mathbb R$$ is our universal set $$X+Y = \mathbb R$$.

Actually proving that for any real $$z$$ there is an $$n$$ so that $$x-\frac 1n \le x \le x+\frac 12$$ can be proven with the Archimedian principal and isn't a given but... I think that's beyond the scope of the problem.

You are correct about D) but you should give a reason. This is simple. If $$z$$ is a real number than there is an integer $$n: n > z$$ then $$z-n < 0 < 100$$ and so $$z-n \in X$$ and $$n \in Y$$ and $$z = (z-n) + n \in X+Y$$ so $$\mathbb R \subset X+Y$$.

C) is however correct.

There are an infinite number of primes so for an $$z \in \mathbb R$$ there is a prime $$p$$ so that $$p > z$$. The $$z - p < 0 < 100$$ so $$z-p \in X$$ and $$p\in Y$$ and $$z = (z-p) + p = z\in X+Y$$ so $$\mathbb R \subset X+Y$$.

• I think you mean $c$ is incorrect, as question asks, for which $X$ and $Y$, $X+Y ≠ R$. – Mathsaddict Aug 14 at 5:56