Assuming an endless universe, there are as many Thursdays as there are days.

I must negate. There are 1/7 as many Thursdays in an infinite amount of time as there are days in general. My opponent brings up bijection. Bijection applies when for every member in one set, there is exactly one member in the other set. This was indeed used to prove that there were just as many nit squares as whole numbers, since each whole number has its own unit square. how do we know that these two sets have the same number of members? Because each whole number must have its own square. The existence of 3 means it is necessary that 3^2 must also exist, since a number must be able to be squared. Now, let us look at what my opponent has attempted to do. He has made two sets, each of which is undeniably infinite given our assumptions [an infinite universe where days continue to hold their same meaning for all time]. Both sets are indeed infinite, but they are also unequal. You see, how do we know that there exists a corresponding Thursday for each day? True, we could do what my opponent has done, and simply write the first few down and make arrows corresponding them. However, this is begging the question. My opponent has simply ASSUMED that there is an equivalent number of members in each set. For the example of numbers and their squares, there was a reasoning behind them having the same number of members. the very existence of any number x necessitates the existence of x^2. To say that there were more numbers than squares would be to say that some numbers did not have corresponding squares, but every whole number has a square as a rule. However, it may be the case that every day does not have its own corresponding Thursday. This unproven point, which is basically the same point he is trying to prove, is something my opponent has just assumed. Now, I will prove that there are 7 times as many days as Thursdays. Both of our sets, number of days passed and number of Thursdays passed, depend on the variable time. I will denote time as t, which will be measured in days. Increasing t by 1 will always mean one more day. Increasing t by 7 will always mean one more Thursday. Let me write this in math notation. The number of days can be represented by the function D(t) = t The number of Thursdays can be represented by the function T(t) = t/7 Where t=1 is the first day we're counting, which in this case is a Friday. Already we can see a one-to-seven correspondence. Now, the functions as we go into infinite. Number of days: lim t->(inf) (t) Number of Thursdays: lim t->(inf) (t/7) Each of these approaches infinity, but we already knew that. To get a ratio, we simply divide them. Since the units are the same, we can combine the two limits into one. Number of thursdays / Number of days: lim t->(inf) ((t/7)/(t)) This gets us inf/inf, which is indeterminate form. We could either cancel out the t's or use L'Hopital's rule, taking the derivatives of the top and bottom functions. That's what I'll do. Top: t/7 Derivative of top: 1/7 Bottom: t Derivative of bottom: 1 Now we are left with this: lim t->(inf) ((1/7)/1) = 1/7 Thus, the ratio of Thursdays to days is 1/7. That is because each Thursday has exactly seven days that correspond to it. Unit squares, however, have a one-to-one correspondence, because the existence of any one whole number X necessitates the existence of it square, and the square of a number necessitates the existence of that number. Each Thursday, as I have shown, necessitates the existence of seven days, and each consecutive seven days necessitates the existence of one Thursday. So while both sets are indeed infinite, one contains exactly 7 times as many members as the other. Your turn.