1. ## 2 exercises

1) {$\displaystyle r_{n}$}=Q , E={x$\displaystyle \in$X : g(x)<f(x)}
Show that E=$\displaystyle \bigcup_{r_{n}}$[{x$\displaystyle \in$X : g(x)<$\displaystyle r_{n}$}$\displaystyle \bigcap${x$\displaystyle \in$X : f(x)>$\displaystyle r_{n}$}]

2) {x$\displaystyle \in$X : g(x)+f(x)<a} = {x$\displaystyle \in$X : g(x)<a-f(x)}.
If f(x) is a countable function show that a-f(x) is countable function too.

2. Originally Posted by SENTINEL4
1) {$\displaystyle r_{n}$}=Q , E={x$\displaystyle \in$X : g(x)<f(x)}
Show that E=$\displaystyle \bigcup_{r_{n}}$[{x$\displaystyle \in$X : g(x)<$\displaystyle r_{n}$}$\displaystyle \bigcap${x$\displaystyle \in$X : f(x)>$\displaystyle r_{n}$}]

2) {x$\displaystyle \in$X : g(x)+f(x)<a} = {x$\displaystyle \in$X : g(x)<a-f(x)}.
If f(x) is a countable function show that a-f(x) is countable function too.
Before spending time on this please make sure this is what you mean.
$\displaystyle E = \bigcup\limits_n {\left[ {\left\{ {x \in X:g(x) < r_n } \right\} \cap \left\{ {x \in X:f(x) > r_n } \right\}} \right]}$

3. Also, what do you mean by "countable function"? I know what "countable sets" are but not "countable function". It may be a translation problem. Do you mean "measurable function".

4. @Plato
Under the $\displaystyle \bigcup$ is $\displaystyle r_{n}$ not n.

@HallsofIvy
It means measurable not countable.

5. $\displaystyle y \in E\; \Rightarrow \;g(y) < f(y)\; \Rightarrow \;\left( {\exists j} \right)\left[ {r_j \in \mathbb{Q} \wedge g(y) < r_j < f(y)} \right]$
$\displaystyle \; \Rightarrow \;y \in \left\{ {x \in X:g(x) < r_j } \right\} \cap \left\{ {x \in X:f(x) > r_j } \right\}$

$\displaystyle \begin{gathered} z \in \bigcup\limits_n {\left[ {\left\{ {x \in X:g(x) < r_n } \right\} \cap \left\{ {x \in X:f(x) > r_n } \right\}} \right]} \hfill \\ \left( {\exists k} \right)\left[ {z \in \left\{ {x \in X:g(x) < r_k } \right\} \cap \left\{ {x \in X:f(x) > r_k } \right\}} \right] \hfill \\ \; \Rightarrow \;g(z) < r_k < f(z)\; \Rightarrow \;z \in E \hfill \\ \end{gathered}$