Let S be a dense subset of R and f : ℝ → ℝ a given function. Define g : S → ℝ by g(x) = f(x). Which of the following statements is necessarily true? 

Concept:

If a function is continuous on a dense set \( S \), it doesn't necessarily imply that the function is

continuous on all of \(\mathbb{R}\) , especially on \( R \setminus S \), the complement of \( S\)  in \(\mathbb{R} \) .

Explanation:

Option 1: Continuity on a dense subset \(S\) does not imply continuity on the whole set \(R\).

A function can be continuous on a dense subset but exhibit discontinuities on \(R \setminus S \).

Therefore, this option is incorrect.

Option 2: \(g\) is defined only on \(S\), so even if \(g\) is continuous on \(S\), it says nothing about \( f \)'s

continuity on the rest of \(​​R\). Continuity of \(g\) does not guarantee the continuity of \( f \) everywhere.

Hence, this option is incorrect.

Option 3:  If \(g(x) = f(x) = 0\) for all \(x \in S \) (which is dense in \(R\)), and \( f \) is continuous on \( R \setminus S \),

by the density of \(S\), \( f \) must be 0 everywhere on \(R\), because a continuous function on a dense set that is

0 must be 0 on the entire set. Therefore, this option is correct.

Option 4: \(g\) being identically 0 on \(S \) and \( f \) being continuous on \( S \)) does not imply \( f \) is identically 0

on \( R \setminus S \). Continuity on \(S\) does not extend to the whole set without further conditions.

Therefore, this option is incorrect.

The correct answer is Option 3).