Simple question but seemingly complicated functions

Question

I am trying to show that if $f, g :[0,1] \rightarrow \mathbb{R} $ then $f(x)+g(x) < t$ if and only if there is a rational number $r $ such that $f(x)<r$ and $g(x)<t-r $. Clearly the if statement is clearly. But if $f(x)+g(x)<t$ how can we say there is a rational number $r $ such that $f(x)<r$ and $g(x)<t-r $. It’s seemingly so simple but yeah it’s confusing me.

I am trying to show that $$\{ x \in [0,1] : f(x)+g(x)<t \}=\bigcup_{r \in \mathbb{Q}}( \{x \in [0,1] : f(x)<r \} \cap \{x \in [0,1] : g(x)<t-r \})$$ but the obviousness seems to be eluding me.

Answer 1

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for any $a,b \in \mathbb{R}$, $a<b$, $[a,b] \cap \mathbb{Q} \neq \emptyset$. Apply this to $a=f(x)$ and $b=t-g(x)$.

Answer 2

The density of the rationals allows this for example you cannot do that for the natural numbers, we want all $x$ such that $\{f(x) + g(x) < t\}$ so let $x$ be one of these $x's$ then look into $f(x) < t - g(x)$ from the density of the rationals there is a rational number $r$ such that $f(x) < r < t-g(x)$ but this gives you $f(x)<r$ and $g(x)<t-r$. Since $x$ was arbitrary so it holds for all such $x$.