The following two images are the ideal low pass filter in the frequency domain. As you can see, the origin (low frequency component), can pass through this filter while the high frequency are blocked. The white value coresponds to 1, the black value coresponds to 0. 
And corespondingly, its spatial domain representation is (2d sinc function):
However, there is a thing that said: "for the spatial domain, the radius of centre component and number of cycle per unit distance from the origin are inversely proportional to the cutoff frequency of the ideal low pass filter".
So when you increase the cutoff frequency, you skretch in frequency domain, but you compress in spatial domain; how does the "number of cycle per unit distance from the origin" decrease? It should increase.
Best Answer
If you scale a function by a factor $a$ in the spatial domain (i.e. the coordinates, or input arguments are scaled), the Fourier transform of the function, i.e. the function in the frequency domain, is scaled by a factor $a^{-1}$. So if you make the cutoff-frequency twice as large, which corresponds to scaling the disc (the function in the frequency domain) by a factor 2, the function in the spatial domain is scaled by a factor ½, which confirms what is said about the radius of centre component.
However, the "number of cycle per unit distance" seems to be a frequency, in which case it should be proportional to the cutoff frequency, not inversely proportional to it. so I agree with you that it should increase and not decrease. Probably just a mistake by the one who wrote that piece of text.
By the way, the two-dimensional analogue of the sinc function is called the sombrero function.