If $f (x) \leqslant g (y)$ for all $x,y\in [0,1]$, then $\sup R(f) \leqslant \inf R(g)$? True/False

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Let $f,g\colon [0,1] \longrightarrow[0,1]$ be function. Let $R (f )$ and $R( g)$ be the ranges of $f$ and $g$, respectively. Which of the following statements is (are) true?

(A) If $f (x) \leqslant g (x)$ for all $x \in [0,1]$, then $\sup R(f) \leqslant \inf R(g)$

(B) If $f (x) \leqslant g (x)$ for some $x \in [0,1]$, then $\inf R(f) \leqslant \sup R(g)$

(C) If $f (x) \leqslant g (y)$ for some $x,y \in [0,1]$, then $\inf R(f)\leqslant\sup R(g)$

(D) If $f (x) \leqslant g (y)$ for all $x,y \in [0,1]$, then $\sup R(f) \leqslant \inf R(g)$

My attempts : if i take $f(x) < g(x)$ see my graphenter image description here

From my graph i can claim that option 1 is not true and option B is true.....i don't know the about option C and D ..im confused pliz help me...

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(A) is false. Your graph is a good starting point.

(B) is true, because $\inf R(f)\leqslant f(x)\leqslant g(x)\leqslant\sup R(g)$.

(C) is true, because $\inf R(f)\leqslant f(x)\leqslant g(y)\leqslant\sup R(g)$.

(D) is true, because every element of $R(f)$ is $\leqslant$ than every element of $R(g)$ and therefore $\sup R(f)\leqslant\inf R(g)$.