If `A sub B "and " C, " then " A in C ` is this statement true ?

Explanation

To determine whether the statement "If A is a subset of B and B belongs to C, then A belongs to C" is true or false, we can analyze the definitions of subsets and set membership step by step.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- A set \( A \) is a subset of a set \( B \) (denoted as \( A \subseteq B \)) if every element of \( A \) is also an element of \( B \).
- An element \( B \) belongs to a set \( C \) (denoted as \( B \in C \)) if \( B \) is one of the elements contained in the set \( C \).

2. **Given Information**:
- We are given that \( A \subseteq B \) and \( B \in C \).

3. **Example Construction**:
- Let's consider an example where \( A = \{2\} \), \( B = \{1, 2\} \), and \( C = \{0, 1, 2\} \).
- Here, \( A \subseteq B \) because the only element of \( A \) (which is 2) is also in \( B \).
- Also, \( B \in C \) because \( B \) (which is the set \( \{1, 2\} \)) is indeed one of the elements in \( C \).

4. **Checking if \( A \in C \)**:
- Now we need to check if \( A \) belongs to \( C \).
- The set \( C \) contains the elements \( 0, 1, 2 \), but it does not contain the set \( A = \{2\} \) as an element.
- Therefore, \( A \notin C \).

5. **Conclusion**:
- Since we have found a case where \( A \subseteq B \) and \( B \in C \) but \( A \notin C \), we conclude that the statement "If \( A \subseteq B \) and \( B \in C \), then \( A \in C \)" is **false**.

### Final Answer:
The statement is **false**.