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1 Set theory

Let \(\Omega = \{ 1,2,3,4\} \) be a set. We can say that

\begin{equation*} 3 \in \{ 1,2,3,4\} \end{equation*}

which means 3 is an element of \(\Omega \). We can’t, however, say that \(3\) is a subset of \(\Omega \) because \(3\) is not a set but an element. On the contrary we say that the set \(\{3\}\) is a subset of \(\Omega \).

Now consider a set \(\Omega = \{ 1,\{ 1\} ,\{ 2,3\} \} \), i.e. there are three elements in \(\Omega \), one number and two sets.

• for \(x=1, 2, ..., n\) we do
• then we remove \(xx-5\)
- – second row1
- – second row2

We can say that

\begin{equation} \label {eq:someq} \{ 2,3\} \in \{ 1,\{ 1\} ,\{ 2,3\} \} f''(x)=7, \end{equation}

because indeed ( 1.1 ) the set \(\{ 2,3\}\) is an element of \(\Omega \) but we cannot say that \(\{ 2,3\} \) is a subset of \(\Omega \).

We can, however, say that \(\{ \{ 2,3\} \} \subset \Omega \) and also \(\{ \{ 2,3\} ,\{ 1\} \} \subset \Omega \), i.e. we are saying that set of subsets is a subset of \(\Omega \).

The rule: in order to be able to say that something is a ¹ subset of a given set or not we must ensure that there are equal levels of brackets in the comparison.

Finally we should also bear in mind that

\begin{eqnarray*} \{ 1,2\} &=& \{ 1\} \cup \{ 2\} \\ \{ \{ 1\} ,\{ 2\} \} &=& \{ \{ 1\} \} \cup \{ \{ 2\} \} \\ \{ \{ 1\} ,\{ 2\} \} &\ne & \{ \{ 1\} \cup \{ 2\} \}. \end{eqnarray*}

Here is another picture.

print("hello x")

def myFunction(x):
    print(xx)
    ahoj
    nazdar
    ciao!!!

import numpy as np

def incmatrix(genl1,genl2):
    m = len(genl1)
    n = len(genl2)
    M = None #to become the incidence matrix
    VT = np.zeros((n*m,1), int) #dummy variable

    #compute the bitwise xor matrix
    M1 = bitxormatrix(genl1)
    M2 = np.triu(bitxormatrix(genl2),1)

    for i in range(m-1):
        for j in range(i+1, m):
            [r,c] = np.where(M2 == M1[i,j])
            for k in range(len(r)):
                 VT[(i)*n + r[k]] = 1;
                 VT[(i)*n + c[k]] = 1;
                 VT[(j)*n + r[k]] = 1;
                 VT[(j)*n + c[k]] = 1;

                 if M is None:
                     M = np.copy(VT)
                 else:
                     M = np.concatenate((M, VT), 1)

                VT = np.zeros((n*m,1), int)

    return M

Therefore if we have a set of subsets \(\{ \{ 1\} ,\{ 2\} ,\{ 3\} \} \) the set e.g. \(\{ 1,2,3\} \) is not an element of the set.

xxx yyy

Below is a table positioned exactly here:

Col1	Col2	Col2	Col3
1	6	87837	787
2	7	78	5415
3	545	778	7507
4	545	18744	7560
5	88	788	6344

¹ test of a footnote with formula ngo \(x+1\)