The shape attribute for numpy arrays returns the dimensions of the array. If Y has n rows and m columns, then Y.shape is (n,m). So Y.shape[0] is n.

In [46]: Y = np.arange(12).reshape(3,4)

In [47]: Y
Out[47]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

In [48]: Y.shape
Out[48]: (3, 4)

In [49]: Y.shape[0]
Out[49]: 3

As far as I know it's just a redundant way to express an empty array. It doesn't seems to matter for python if you have rows of "emptiness".

Let's say we have a give array a:

import numpy as np

a = np.zeros((0,100))

If we print a all we get is the empty array itself:

print(a)

>>> []

Moreover we can actually see that despite this a maintain it's shape"

np.shape(a)

>>> (0, 100)

But if you try to access a given element by position, e.g:

print(a[0])

or

print(a[0][0])

You get an IndexError :

IndexError: index 0 is out of bounds for axis 0 with size 0

Therefore I believe that the mathematical meaning of the empty arrays, despite the shape you assign to them, is the same.