If `A sube B and C` be any set , then prove that:
`A xx C sube B xx C`

Explanation

To prove that \( A \times C \subseteq B \times C \) given that \( A \subseteq B \) and \( C \) is any set, we can follow these steps:

### Step 1: Understand the Cartesian Product
The Cartesian product \( A \times C \) is defined as the set of all ordered pairs \( (a, c) \) where \( a \in A \) and \( c \in C \). Similarly, \( B \times C \) is the set of all ordered pairs \( (b, c) \) where \( b \in B \) and \( c \in C \).

### Step 2: Take an arbitrary element from \( A \times C \)
Let \( (a, c) \) be an arbitrary element of \( A \times C \). By the definition of the Cartesian product, this means that \( a \in A \) and \( c \in C \).

### Step 3: Use the subset property
Since we know that \( A \subseteq B \), it follows that every element \( a \) in \( A \) is also in \( B \). Therefore, we can conclude that \( a \in B \).

### Step 4: Form the ordered pair for \( B \times C \)
Now that we have \( a \in B \) and \( c \in C \), we can form the ordered pair \( (a, c) \). Since \( a \) is an element of \( B \) and \( c \) is an element of \( C \), it follows that \( (a, c) \) is also an element of \( B \times C \).

### Step 5: Conclude the proof
Since our choice of \( (a, c) \) was arbitrary, we have shown that every element of \( A \times C \) is also an element of \( B \times C \). Thus, we can conclude that \( A \times C \subseteq B \times C \).

### Final Statement
Therefore, we have proved that if \( A \subseteq B \), then \( A \times C \subseteq B \times C \).

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