 It can be shown that every well-ordered series has an ordinal number…and that the series of all ordinals (in order of magnitude) is well-ordered. It follows that the series of all ordinals has an ordinal number, \(\Omega\) say. But in that case the series of all ordinals including \(\Omega\) has the ordinal number \(\Omega + 1\), which must be greater than \(\Omega\), Hence \(\Omega\) is not the ordinal number of all ordinals. (PM Vol. I, 1925: 61 and 1910: 63)