A vertex of a triangle belongs to the triangle. A triangle belongs to the set of all triangles. But, a vertex is not itself a triangle.


Let $y=\{1\}$. And $x=\{y\}$. Then $1\in y\land y\in x$, but $1\not\in x$.


The difference between $\subset$ and $\in$ is that the former applies to expressions at the same level of nesting and the latter applies to expressions at one level of nesting apart from each other. So when you chain two $\in$'s together you get something at two levels of nesting, which is not in general comparable to a single $\in$. On the other hand, since ...


$42 \in \mathrm{Even} \in \mathcal{P}(\mathbb{Z})$ but $42 \not\in \mathcal{P}(\mathbb{Z})$ because 42 is not a set of integers. $\text{Peter} \in \text{Humans} \in \text{Species}$ but $\text{Peter} \not\in \text{Species}$ because Peter is not a species.


Consider the empty set $\phi,$ which has no members. And $x=\{\phi\}$ has one member (namely, $\phi$ is the only member of $x$). And let $y=\{x\}.$ So $\phi \in x$ and $x\in y.$ But $\phi\not\in y,$ because the only member of $y$ is $x,....$ and $x$ is not $\phi$ because $x$ has a member while $\phi$ has none.


Belonging is not transitive because we don't want it to be. Suppose I have sets $A = \{1, 2\}$ and $B = \{3, 4\}$. Now imagine that we write, "Let $C = \{A, B\}$." When we say "Let $C = \{A, B\}$," what we're saying is that we want $C$ to be a set with exactly two elements: one of the elements is $A$, and the other element is $B$. If we wanted $C$ to have ...


Sure, the binary relation $R = \{(1,2), (3,4)\}$ on the set $A = \{1,2,3,4\}$ is transitive. Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) \in R$ and $(y,z) \in R$, there is nothing to check.


My dog belongs to me and I belong to the American Mathematical Society... and so....


Yes, your relation is indeed transitive. The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter. Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation. Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place)...


Based on your response to other answers, your question seems to be "Why do we define the belonging relation ($\in$) in a way that cares about the level of nesting of sets?", i.e. why do we say that $a \notin \{\{a\}\}$? We could define a relation that ignores nesting, a sort of recursive belonging that works the way you seem to want belonging to work (with $...


A relation $R$ fails to be transitive only if you can find $a,b,c$ such that $(a,b)\in R$ and $(b,c)\in R$ are true but $(a,c)\in R$ is not true. Your relation is transitive since you can not find such $a,b,c$ to contradict transitivity.


Hint: Use the definitions. For example, $R$ is antisymmetric, if ($aRb$ and $bRa$) implies $a=b$. Here $aRb$ is a shorthand for $(a,b)\in R$, thus $(a,b)\in R^{-1}$ means $bRa$, and $a=b$ means exactly that $(a,b)\in I$.


Or one can determine the truth value of the proposition : $\forall x,y,z \in X : ( x \rm{R} y \land y\rm{R} z )$ $\implies$ $x\rm{R}z $. Let P= $( x \rm{R} y \land y\rm{R} z )$ , Q= $x\rm{R}z$ Suppose we have $ x\rm{R}y$, so its value is true, the truth value of $ y\rm{R}z$ is always false. So the truth value of $P$ is false, based on the truth ...


HINT. Formally, is described the model $$z(x,y) = e^{a+bx+cy},$$ or $$\ln z = a+bx+cy,\tag1$$ where $b<0, c>0,$ $$x_i=12+2.25i,\quad y_i= 100 + 24j.$$ The table data model is $$z_{i,j} = e^{\large \frac xy+1}\tag2$$ Assuming the discrepancy in the form of $$d(a,b,c) = \sum w_{i,j}(\ln z_{i,j} - a - bx_i - Cy_j)^2,\tag3$$ where $w_{ij}$ is the ...


An exponential function is kind of $f(x)=z=a+b·e^{c·x}$ First compute all regresions, one for each row. For example, the row $y=21.00$ results in $z=2.89657+1.73346\cdot e^{(-0.0134048\cdot x)}$ And the row $y=23.25$ results in $z=2.91707+1.96536\cdot e^{(-0.135084\cdot x)}$ Next calculate the z values for given $x=150$ for all rows. In the example $z(21....

Only top voted, non community-wiki answers of a minimum length are eligible