Suppose that I have an infinite set, where the definition of infinite we are using is that contains a countable (i.e. bijective to the naturals) subset. This definition assumes the axiom of choice (don't worry about it), but if you'd like, replace the general set in the rest of this post by the real numbers, which we know contain the natural numbers as a countable subset.
Let X be an infinite set, and {x_0,x_1,x_2,....} be a countable subset. Suppose we want to add a point to x. The context that this came up in was constructing a bijection from the half- closed ray [-2,inf) to the real numbers.
You might know that if this were an open interval (-2,inf) then a bijection would easily be given by R->(-2,inf), x|-> (e^x)-2. Its inverse is log(x+2). We want to "hide" the point, get rid of it somehow.
In other words, we add a point * to our set which is disjoint from it, and we want to construct a bijection X u {*} -> X, thereby hiding the point. In the case of the problem, this is a bijection [-2,inf)->(-2,inf), and our point * would be -2.
You might know of Hilbert's Hotel, which has infinitely many rooms numbered 1,2,3,.... and is fully occupied. A new guest requests a room, so to accomodate them, the guest at room 1 moves to room 2, the guest at room 2 to room 3, and so on, so all the guests still have a room, but now we have room 1 for our new guest! We employ the same idea here, defining a function by:
In the case of our bijection [-2,inf)->(2,inf), we use the natural numbers, so we map -2 to 0, 0 to 1, 1 to 2 etc, and all other points to themselves.
Now we can compose our bijection with the bijection (-2,inf)->R we described earlier to show that [-2,inf) is in bijection with R.
