To prove that there exists a one-one correspondence (bijective function) between S x T and T x S, we need to construct a function that maps each element in S x T to a unique element in T x S, and vice versa.

Let's define a function f: S x T - T x S as follows:

For any element (s, t) in S x T, let f(s, t) = (t, s).

We need to show that this function is both injective (one-to-one) and surjective (onto).

1. Injective: Assume we have two pairs (s1, t1) and (s2, t2) in S x T such that f(s1, t1) = f(s2, t2). This means that (t1, s1) = (t2, s2). From this, we can conclude that t1 = t2 and s1 = s2. Therefore, (s1, t1) = (s2, t2), which shows that f is injective.

2. Surjective: For any element (t, s) in T x S, we can find an element (s, t) in S x T such that f(s, t) = (t, s). This shows that f is surjective.

Since f is both injective and surjective, it is bijective. Therefore, there exists a one-one correspondence between S x T and T x S.

Question

To prove that there exists a one-one correspondence (bijective function) between S x T and T x S, we need to construct a function that maps each element in S x T to a unique element in T x S, and vice versa.

Let's define a function f: S x T -> T x S as follows:

For any element (s, t) in S x T, let f(s, t) = (t, s).

We need to show that this function is both injective (one-to-one) and surjective (onto).

1. Injective: Assume we have two pairs (s1, t1) and (s2, t2) in S x T such that f(s1, t1) = f(s2, t2). This means that (t1, s1) = (t2, s2). From this, we can conclude that t1 = t2 and s1 = s2. Therefore, (s1, t1) = (s2, t2), which shows that f is injective.

2. Surjective: For any element (t, s) in T x S, we can find an element (s, t) in S x T such that f(s, t) = (t, s). This shows that f is surjective.

Since f is both injective and surjective, it is bijective. Therefore, there exists a one-one correspondence between S x T and T x S.

Knowee AI · Accepted Answer

To prove that there exists a one-one correspondence (bijective function) between S x T and T x S, we need to construct a function that maps each element in S x T to a unique element in T x S, and vice versa.

Let's define a function f: S x T -> T x S as follows:

For any element (s, t) in S x T, let f(s, t) = (t, s).

We need to show that this function is both injective (one-to-one) and surjective (onto).

1. Injective: Assume we have two pairs (s1, t1) and (s2, t2) in S x T such that f(s1, t1) = f(s2, t2). This means that (t1, s1) = (t2, s2). From this, we can conclude that t1 = t2 and s1 = s2. Therefore, (s1, t1) = (s2, t2), which shows that f is injective.

2. Surjective: For any element (t, s) in T x S, we can find an element (s, t) in S x T such that f(s, t) = (t, s). This shows that f is surjective.

Since f is both injective and surjective, it is bijective. Therefore, there exists a one-one correspondence between S x T and T x S.

If S and T are non-empty sets. Prove that there exists a one-one correspondence(bijective) between 𝑆 × 𝑇 and 𝑇 × 𝑆

Question

Solution

Similar Questions

Upgrade your grade with Knowee