To show that T is countable, we need to provide a bijection f: N → T. A bijection is a function that is both injective (one-to-one) and surjective (onto).

Let's define the function f: N → T such that f(n) = t(n) for all n in N.

Injectivity:
f is injective if for every m, n in N, if f(m) = f(n) then m = n.
In our case, if f(m) = f(n) then t(m) = t(n). But since the t(n) are an enumeration of T, no two different n can have the same t(n). So, m must equal n. Hence, f is injective.

Surjectivity:
f is surjective if for every t in T, there is an n in N such that f(n) = t.
In our case, since the t(n) are an enumeration of T, for every t in T, there is an n in N such that t(n) = t. Hence, f is surjective.

Therefore, f is a bijection from N to T, so T is countable.

Question

To show that T is countable, we need to provide a bijection f: N → T. A bijection is a function that is both injective (one-to-one) and surjective (onto).

Let's define the function f: N → T such that f(n) = t(n) for all n in N.

Injectivity: 
f is injective if for every m, n in N, if f(m) = f(n) then m = n. 
In our case, if f(m) = f(n) then t(m) = t(n). But since the t(n) are an enumeration of T, no two different n can have the same t(n). So, m must equal n. Hence, f is injective.

Surjectivity: 
f is surjective if for every t in T, there is an n in N such that f(n) = t. 
In our case, since the t(n) are an enumeration of T, for every t in T, there is an n in N such that t(n) = t. Hence, f is surjective.

Therefore, f is a bijection from N to T, so T is countable.

Knowee AI · Accepted Answer

To show that T is countable, we need to provide a bijection f: N → T. A bijection is a function that is both injective (one-to-one) and surjective (onto).

Let's define the function f: N → T such that f(n) = t(n) for all n in N.

Injectivity: 
f is injective if for every m, n in N, if f(m) = f(n) then m = n. 
In our case, if f(m) = f(n) then t(m) = t(n). But since the t(n) are an enumeration of T, no two different n can have the same t(n). So, m must equal n. Hence, f is injective.

Surjectivity: 
f is surjective if for every t in T, there is an n in N such that f(n) = t. 
In our case, since the t(n) are an enumeration of T, for every t in T, there is an n in N such that t(n) = t. Hence, f is surjective.

Therefore, f is a bijection from N to T, so T is countable.

Suppose T is an inﬁnite set and t(1), t(2), . . . , t(n), . . . is an enu-meration of T . Show that T is countable by providing a bijection f : N →T , brieﬂy explain why f is indeed injective and surjective

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