# Finding support of a function

Let $$X$$ be a set and $$f$$ be a function from $$X$$ to $$\{0, 1\}$$, the field with two elements. The support of $$f$$ is the set $$f ^ {-1}$$ (1), which we denote by $$\DeclareMathOperator{\supp}{supp}\supp(f)$$. For a collection $$F$$ of such functions define $$\:\supp(F) = \bigcap_{f∈F} \supp(f)$$

Note that when $$F = ∅, \:\supp(F) = X$$, and that for such functions, $$f$$ is the characteristic function of its support, i.e., $$f = χ_{\supp(f)}$$. Find the supports of $$f · g$$, $$1 − f$$, $$f + g − f · g$$, and $$1 − f + f · g$$ in terms of the supports of $$f$$ and $$g$$.

Do now know where to start. Will be helpful to get at least the first one to see how it works.

Let $$f \cdot g \colon X \to \{0,1\}$$ be the function defined by $$(f \cdot g) (x) = f(x) \cdot g(x)$$ for all $$x \in X$$. The support of $$f \cdot g$$ is the intersection of the supports of $$f$$ and $$g$$. Indeed, by definition of support, \begin{align} \operatorname{supp}(f \cdot g) &= \{x \in X \mid f(x) \cdot g(x) = 1\} \\ &= \{x \in X \mid f(x) = 1 \text{ and } g(x) = 1\} \\ &= \{x \in X \mid f(x) = 1\} \cap \{x \in X \mid g(x) = 1\} = \operatorname{supp}(f) \cap \operatorname{supp}(g). \end{align}
Let $$1 \!-\! f \colon X \to \{0,1\}$$ be the function defined by $$(1 \!-\! f)(x) = 1 - f(x)$$ for all $$x \in X$$. The support of $$1 \!-\!f$$ is the complement of the support of $$f$$. Indeed, by definition of support, \begin{align} \operatorname{supp}(1 \!-\! f) &= \{x \in X \mid f(x) = 0\} \\ &= \{x \in X \mid f(x) \neq 1\} \\ &= X \smallsetminus \{x \in X \mid f(x) = 1\} \ = X \smallsetminus \operatorname{supp}(f). \end{align}
• @Amalya - Yes, you're right. What about $\text{supp}(f + g - f \!\cdot\! g)$? – Taroccoesbrocco Feb 16 at 14:35