Something most people understand, but many probably cannot prove on their own. The proof follows, but first, a little background:

Two sets are considered equinumerous if they have the same number of elements. for example, the set {2 5 6} is equinumerous with the set {-512 5 345}. Also, the set of all natural numbers {1 2 3 4 5...} has the same number of elements as the set of all even positive integers {2 4 6 8 10...}; they both have a cardinality of aleph null, meaning each has aleph null elements. To prove that sets with a transfinite number of elements have the same number of elements, a bijective, or one-to-one function from one set to the other must exist. So, we can prove that there are as many even postive integers as there are natural numbers by finding the bijective function between the two, and that function is f(x) = 2x.

So, to prove the statement at the title of this node, we have to find a bijective function mapping the numbers in the interval (0,1) onto the set of all real numbers, from negative infinity to positive infinity. The first such function I could find was a piecewise function defined as f(x) = (ln(x)-ln(1/2)) when x is less than 1/2 and f(x) = -ln(1-x) + ln(1/2) when x is greater than 1/2. Since a bijective function between (0,1) and the set of all real numbers exists, the two sets are equinumerous, and each contains the same number of elements.

A smoother one is the tan function, which of course goes from negative to positive infinity in the compass of -pi/2 to pi/2, so just compose it with a function that scales the interval (0, 1) up.

Frustratingly, it is obvious that taking the closed interval [0, 1] trivially adds two points to a set of cardinality c, yet extending the bijection to include them is nontrivial.

Thanks to ariels for pointing out it's in fact impossible to have such a continuous bijection, since it would be from a compact set to an open one.

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