Verified answer

Here's a sketch:

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First, note that S ~ [0, 2π), because we have the bijection

f: [0, 2π) → S defined by f(t) = (cos t, sin t).

Since [0, 2π) is an interval, we have [0, 2π) ~ R. Hence, S ~ R.

Next, note that since [0, ∞) ~ R, we have T = [0, ∞) x R ~ R x R = R^2.

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So, it suffices to show that |R| = |R^2|.

Since (0, 1) ~ R, via the bijection f(x) = tan(π(x - 1/2)),

it suffices to show that |(0, 1)| = |(0, 1) x (0, 1)|.

We create a two 1-1 maps:

(i) f : (0, 1) x (0, 1) → (0, 1) defined by f(0.a(1)a(2)..., 0.b(1)b(2)...) = 0. a(1) b(1) a(2) b(2) ...

[The above is in decimal notation; so we are 'intertwining' decimals together.

We need to rule out repeating 9's for a unique decimal representations in (0, 1) x (0, 1).]

(ii) g : (0, 1) → (0, 1) x (0, 1) defined by f(x) = (x, x).

(That this is a bijection is immediate.)

Therefore, |(0, 1)| = |(0, 1) x (0, 1)| by the Cantor-Bernstein-Shroeder Theorem.

I hope this helps!