1. 1) Show that set {(x,y,z) $\in$ $\Re^3$ : z=0) is a zero subset of $\Re^3$

2) Show that set {(x,y,z,t) $\in$ $\Re^4$ : - $\infty$<x<+ $\infty$, - $\infty$<y<+ $\infty$, - $\infty$<z<+ $\infty$, t $\in$Q) is a zero subset of $\Re^4$

I have these two problems in my Measure Theory lesson... If anyone can help i would appreciate it...

I would be happy if someone tell me something to begin with these exercises if not the full solution...
Especially what i am doing with - $\infty$ < x < + $\infty$

2. Originally Posted by SENTINEL4
1) Show that set {(x,y,z) $\in$ $\Re^3$ : z=0) is a zero subset of $\Re^3$

2) Show that set {(x,y,z,t) $\in$ $\Re^4$ : - $\infty$<x<+ $\infty$, - $\infty$<y<+ $\infty$, - $\infty$<z<+ $\infty$, t $\in$Q) is a zero subset of $\Re^4$
1) Given m, n and k, the box $[m,m+1]\times[n,n+1]\times[-1/k,1/k]$ has volume 2/k. Taking the intersection over all natural numbers k, you see that the set $\{(x,y,z)\in\mathbb{R}^3:m\leqslant x\leqslant m+1,\ n\leqslant y\leqslant n+1,\ z=0\}$ has measure 0. Now take the (countable) union over all integers m and n.

2) By the same argument as in 1), the set $\{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ has measure 0, for each rational number r. Now take the union over all such r.

3. In 2), at $\{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ maybe it is t=r instead of z=r ??

4. Originally Posted by SENTINEL4
In 2), at $\{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ maybe it is t=r instead of z=r ??
Yes.

5. Originally Posted by Opalg
1) Given m, n and k, the box $[m,m+1]\times[n,n+1]\times[-1/k,1/k]$ has volume 2/k. Taking the intersection over all natural numbers k, you see that the set $\{(x,y,z)\in\mathbb{R}^3:m\leqslant x\leqslant m+1,\ n\leqslant y\leqslant n+1,\ z=0\}$ has measure 0. Now take the (countable) union over all integers m and n.

2) By the same argument as in 1), the set $\{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ has measure 0, for each rational number r. Now take the union over all such r.
Opalg's eplanation is clear. If I understand Opalg's 1), then by parrallel argument, should 2) be unioned over three indexes?:
2) By the same argument as in 1), the set $\{(x,y,z,t)\in\mathbb{R}^4:t=r\}$ has measure 0, for any rational number r. Now take the (countable) union over all integers m, n and p.