If $A$ and $B$ are sets, then $A$ is called a proper subset of $B$, if, and only if, every element of $A$ is also an element of $B$, and $B$ contains at least one element that is not contained in $A$.

\(A \subset B \Longleftrightarrow \forall \; x_{1}\; (x_{1} \in A \Longrightarrow x_{1} \in B) \; \wedge \; \exists \; x_{2}\; (x_{2} \in B \wedge x_{2} \notin A)\).

Symbol | Meaning |
---|---|

$ \subset $ | is a proper subset of |

$ \Longleftrightarrow $ | if, and only if |

$ \forall $ | for all (universal quantifier) |

$ \in $ | is an element of |

$ \Longrightarrow $ | if, then |

$ \wedge $ | and |

$ \exists $ | it exists (existential quantifier) |

$ \notin $ | is not an element of |

Let $A$ and $B$ be the following finite sets: $A = \{1, 2, 3\}$, $B = \{1, 2, 3, 4, 5\}$. All three elements of $A$ are in $B$, but there are also the elements $4$ and $5$ that are containted in B, but not in $A$. Therefore, $A$ is a proper subset of $B$ and we can write $A \subset B$.

As we have seen in the definition of subsets (see link below), the symbol that expresses “is a subset of” $\subseteq$ contains also some sort of equals sign. Due to the fact that a proper subset $A$ by definition does not contain all the element of set $B$, $A$ being a proper subset of $B$ is also denoted as $A \subsetneq B$.

Starting from he formal definition, a set $A$ is *not* a proper subset of set $B$, if there is at least an element in set $A$ that is not contained in set $B$ or in case every element of $B$ is also contained in set $A$ (see the non-examples section below).

Symbol | Meaning |
---|---|

$ \not\subset $ | is not a proper subset of |

$ \Longleftrightarrow $ | if, and only if |

$ \exists $ | it exists (existential quantifier) |

$ \in $ | is an element of |

$ \wedge $ | and |

$ \notin $ | is not an element of |

$ \vee $ | (not exclusive) or |

$ \forall $ | for all (universal quantifier) |

$ \Longrightarrow $ | if, then |

Let $A$ and $B$ be the following sets: $A = \{1, 2, 3, 4\}$, $B = \{1, 2, 3\}$. In this case, set $A$ contains an element, that is not contained in set $B$, and is therefore not even a subset of $B$. This is an example for the first part of the expression above: $\exists \; x_{1}\; (x_{1} \in A \wedge x_{1} \notin B)$.

Let $A$ and $B$ be the following sets: $A = \{1, 2, 3\}$, $B = \{1, 2, 3\}$. In this case, every element that is contained in set $B$ is also contained in set $A$. But by definition of proper subset, set $B$ must contain at least one element that is not contained in $A$. This is an example for the second part of the expression above: $\forall \; x_{2}\; (x_{2} \in B \Longrightarrow x_{2} \in A).$

It is clear that these two cases cannot occur simultaneously.

Given two boxes, if all the objects of the first box are also contained within the second box, and the second box contains at least one additional object that is not contained the first box, then the first box is said to be a proper subset of the second box.

If $A$ and $B$ are sets, then $A$ is called a subset of $B$, if, and only if, every element of $A$ is also an element of $B$.

Symbol | Meaning |
---|---|

$ \subseteq $ | is a subset of |

$ \Longleftrightarrow $ | if, and only if |

$ \forall $ | for all (universal quantifier) |

$ \in $ | is an element of |

$ \Longrightarrow $ | if, then |

Let $A$ and $B$ be the following (finite) sets: $A = \{1, 3, 5, \{6\}\}$, $B = \{1, 2, 3, 4, 5, \{6\}\}$. All elements of $A$ are the four elments 1, 3, 5, and $\{6\}$. All four elements are in $B$. Therefore, $A$ is indeed a subset of $B$. We can therefore write $A \subseteq B$.

Let’s say that we want to enumerate *all* the possible subsets of a finite set.
The question arises as to how we can do this without running the risk of forgetting a possible subset.
Actually, there is a useful method:
For the sake of simplicity, let’s imagine that the set $C$ has only one element: $ C = \{ \alpha \} $.
Now, by definition of subset, $C = \{\alpha\} $ itself is a possible subset of $C$.
We could also remove the element $\alpha$ from it.
And now, also $ C = \{\}$ is a possible subset.
So basically, for each element in a set, either the element is included in the subset or it’s not.
Therefore, the possible subsets of $C$ are actually two.

Now, let $D$ be the following (finite) set: $ D = \{\alpha, \beta, \{\bigstar\}\}$. As we have seen in the previuous example, each of the three elements is either contained in the subset or not, so $\alpha$ is in $\boxdot$ or out $\square$ (two possibilities), $\beta$ is in $\boxdot$ or out $\square$ (two possibilities), and finally $\{\bigstar\}$ is in $\boxdot$ or out $\square$ (two possibilities), which we can also denote as $2 \times 2 \times 2 = 2^{3} = 8$. If we generalize this, we get the following formula for calculating all the subsets of a set, that is

\[2^{n}\]where $n$ is the number of elements in the (finite) set at hand. As we have seen, by using the method described, the total number of subsets of set $D$ is $8$:

1. | $ \{\} $ |

2. | $ \{\alpha\}$ |

3. | $ \{\beta\}$ |

4. | $ \{ \{ \bigstar \} \}$ |

5. | $ \{ \alpha, \beta \}$ |

6. | $ \{ \alpha, \{ \bigstar \} \}$ |

7. | $ \{ \beta, \{ \bigstar \} \}$ |

8. | $ \{ \alpha, \beta, \{ \bigstar \} \}$ |

Starting from he formal definition of a subset, a set $A$ is *not* a subset of another set $B$, if not every element of $A$ is also contained in set $B$.
That is, if there is at least one element in set $A$, that is not contained in set $B$.

Symbol | Meaning |
---|---|

$ \nsubseteq $ | is not a subset of |

$ \Longleftrightarrow $ | if, and only if |

$ \exists $ | it exists (existential quantifier) |

$ \in $ | is an element of |

$ \wedge $ | and |

$ \notin $ | is not an element of |

Let A and B be the following sets: $A = \{1, 3, 5, 6\}$, $B = \{1, 2, 3, 4, 5\}$. All elements of $A$ are the four elments 1, 3, 5, and 6. But there is at least one element of A, that is not containted in set B: the element 6 in our example. Therefore, A is not a subset of B ($A \nsubseteq B$).

Given two boxes, if all the objects of the first box are contained within the second box, the first box is said to be a subset of the second box. The second box may - but doesn’t need to - contain supplementary objects which are not contained within the first box. That’s why the symbol that expresses “is a subset of” $\subseteq$ contains also some sort of equals sign.

If we imagine the sets as boxes and the subsets as possible configurations of boxes, then it is clear that a box can of course also be empty.

This article presents a very famous proof that the square root of 2 cannot be expressed by a rational number.

The target audience is everyone who is interested in the topic or is curious. The Greek proof that there is no rational number whose square equals 2 is one of the great intellectual achievements of humanity and it should be experienced by every educated person (S. J. Axler, Algebra & trigonometry: with student solutions manual, Hoboken, N.J, 2012, p. 4).

None.

Here you will find the actual article.

]]>Mathematical induction, a deductive process, is known to be a proof method for proving statements about positive integers. This article provides an overview of this proof technique, ranging from the terminological aspects to the actual description of the proof.

Everyone who is interested in the topic or is curious. In case you’re not familiar with the summation notation: read anyway and don’t be intimidated by it. You should still be able to grasp the concept.

None.

Here you will find the actual article.

]]>