2 exercises

• May 12th 2009, 10:46 AM
SENTINEL4
2 exercises
1) { $r_{n}$}=Q , E={x $\in$X : g(x)<f(x)}
Show that E= $\bigcup_{r_{n}}$[{x $\in$X : g(x)< $r_{n}$} $\bigcap${x $\in$X : f(x)> $r_{n}$}]

2) {x $\in$X : g(x)+f(x)<a} = {x $\in$X : g(x)<a-f(x)}.
If f(x) is a countable function show that a-f(x) is countable function too.
• May 12th 2009, 11:03 AM
Plato
Quote:

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

2) {x $\in$X : g(x)+f(x)<a} = {x $\in$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.
$E = \bigcup\limits_n {\left[ {\left\{ {x \in X:g(x) < r_n } \right\} \cap \left\{ {x \in X:f(x) > r_n } \right\}} \right]}$
• May 12th 2009, 11:40 AM
HallsofIvy
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".
• May 12th 2009, 12:21 PM
SENTINEL4
@Plato
Under the $\bigcup$ is $r_{n}$ not n.

@HallsofIvy
It means measurable not countable.
• May 12th 2009, 12:37 PM
Plato
$y \in E\; \Rightarrow \;g(y) < f(y)\; \Rightarrow \;\left( {\exists j} \right)\left[ {r_j \in \mathbb{Q} \wedge g(y) < r_j < f(y)} \right]$
$\; \Rightarrow \;y \in \left\{ {x \in X:g(x) < r_j } \right\} \cap \left\{ {x \in X:f(x) > r_j } \right\}$

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