I'm reading some past papers for my next week Set Theory exams, and I found these two problems:
1).Let $n \in \mathbb{N},$ and for $i \leq n, f_i: \mathbb{R}\rightarrow \mathbb{R}$ functions such that: $(\forall i \leq n)(\forall x \in \mathbb{R})[f_i(x)>0].$ Show that there exist $\varepsilon>0, B\subseteq \mathbb{R}$ with $|B|=|\mathbb{R}|$ such that: $(\forall i \leq n)(\forall x \in B)[f_i(x)>\varepsilon].$
Any hint on how to begin would be appreciated.
First, try $\varepsilon=1$, and let $B_1$ be the set of those $x\in\mathbb R$ where all the functions $f_i$ simultaneously have values $>1$. If $|B_1|=|\mathbb R|$, you're done, so assume $|B_1|<|\mathbb R|$. In that case, try again with $\varepsilon=\frac12$, define $B_{\frac12}$ in analogy with the preceding, and observe that, if $|B_{\frac12}|=|\mathbb R|$, you're done. If not, try again with $\varepsilon=\frac13$, and so forth. If you ever get $|B_{\frac1k}|=|\mathbb R|$, you're done. What if, for all $k$, you get $|B_{\frac1k}|<|\mathbb R|$? Well, notice that $\bigcup_kB_{\frac1k}=\mathbb R$. (This uses the fact that you have only finitely many $f_i$'s.) So you'd have $\mathbb R$ represented as the union of countably many sets $B_{\frac1k}$ of strictly smaller cardinality, which you know is impossible.