Level Sets of a Continuous Function

Detailed Solution

Define $$ g(x)=f(x)-x. $$ Since $f$ is continuous and the function $x\mapsto x$ is continuous, their difference $g$ is also continuous. Now express the three sets in terms of $g$: $$ A=\{x:g(x)>0\}, $$ $$ B=\{x:g(x)=0\}, $$ and $$ C=\{x:g(x)<0\}. $$ For a continuous function, the inverse image of an open set is open and the inverse image of a closed set is closed. Since $(0,\infty)$ and $(-\infty,0)$ are open subsets of $\mathbb R$, $$ A=g^{-1}((0,\infty)) $$ and $$ C=g^{-1}((-\infty,0)) $$ are open sets. Also, $\{0\}$ is a closed subset of $\mathbb R$. Therefore, $$ B=g^{-1}(\{0\}) $$ is closed. Hence, $$ \boxed{A\text{ and }C\text{ are open, while }B\text{ is closed}.} $$ Now check the remaining options. A is false. For example, if $f(x)=x$, then $$ B=\mathbb R,\qquad A=C=\varnothing. $$ Thus all three sets need not be non-empty. C is false. Again, for $f(x)=x$, $$ B=\mathbb R, $$ which is non-empty but is not compact because $\mathbb R$ is unbounded. D is false because although $A$, $B$, and $C$ are pairwise disjoint and $$ A\cup B\cup C=\mathbb R, $$ the set $B$ need not be open. For example, when $f(x)=x$, we have $B=\mathbb R$, which happens to be open, but in general $B$ need not be open. For instance, if $f(x)=x^2$, then $$ B=\{0,1\}, $$ which is closed but not open in $\mathbb R$. Therefore, the only statement that must be true is $$ \boxed{\mathrm{B}}. $$