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8

You have $x\in B$ and $x\not\in B$. Notice a problem here?


6

It seems like you have picked up on the main issue via avid19's push in the right direction, but I would encourage you to think about associativity of $\cap$, something that would immediately answer the question for you: \begin{align} (Α\setminus C)\cap(B\setminus C)\cap(A\setminus B)&= (A\cap C^C)\cap(B\cap C^C)\cap(A\cap B^C)\tag{by defn.}\\[0.5em] ...


6

You can do it in two ways (at least): You can note that $\Bbb{|R|\leq|R\times Z|\leq|R\times R|}$, and apply some general theorems on cardinals. You can note that $\Bbb R$ and $[0,1)$ have a bijection between them, so you're really looking at $[0,1)\times\Bbb Z$ in terms of cardinality. Now use your imagination to write down a bijection with $\Bbb R$.


6

A picture is worth a thousand words. $A-(B \cup C)$ is everything in A that is not also in either B or C. $A-(B \cap C)$ is everything in A that is not also in both B and in C.


5

The set of cats is not a rational number. So $x$ could be the set of cats. That's probably not a real number. In slightly more serious terms, where do these $x$ belong? You need a set to draw these from. Read about dedekind cuts for one approach that derives the reals from subsets of rationals. Cauchy's completeness argument defines them uses sequences of ...


4

Let $x \in B$. We want to prove $x \in \overline{A}$, i.e. that $x \notin A$. So assume $x \in A$. But then, we have $x \in A$ and $x \in B$, which means $x \in A \cap B$. However, by assumption, $A \cap B$ is a empty. This is a contradiction. This proves $x \notin A$. Actually, the converse is also true: $A \cap B$ is empty if and only if $B$ is ...


3

Let's assume $A\cap B =\varnothing$ (start hypothesis) Let $x \in B$ Since $A$ and $B$ are disjoint (start hypothesis), then $x \notin A$ By definition of $\overline A$, since $x \notin A$ then $x \in \overline A ~~~~(= \Omega - A)$ Therefore $B\subseteq \overline A$, because for all $x \in B$, we have $x \in \overline A$ Note that the reciprocal ...


3

A way to do this is to take a denumerable set in $D \subset (0,1)$, to define a bijection as the identity outside $D$, to map the first element of $D$ to $1$ [and the second to $0$] and to "push" the remaining elements of $D$ down by $1$ [or $2$] to get a bijection with $(0,1]$, and $[0,1]$.


3

A counterexample would be sufficient and just writing A=B=C={a} would disprove the statement.


3

Once you settle on what are the rationals and what are the real numbers, it becomes particularly easy, since $\sf ZFC$ proves that if $A$ and $B$ are sets, then $A\setminus B$ is a set. Why? Apply the axiom of separation for $\varphi(x,p_1)$ (where $p_1$ is a parameter) taking the formula to be $x\notin p_1$. Now the class $\{x\in A\mid x\notin B\}$ is a ...


3

Hint: Use two steps and show $$(R\setminus S)\setminus T\subseteq R\setminus S\subseteq R\setminus(S\setminus T)$$


3

If $p$ is false then $p\rightarrow q$ is true for any statement $q$. This becomes evident by realizing that $p\rightarrow q$ is actually the same as $\neg p\vee q$. So defining $p$ as the (false) statement $x\in E\cap E^c$ and $q$ as the statement $x\in\varnothing$ we arrive at the conclusion that $x\in E\cap E^c\rightarrow x\in\varnothing$ is a true ...


3

You did prove that the implication $x \in E^c \cap E \implies x \in \varnothing$ is true, which means that $E^c\cap E \subseteq \varnothing$, but there's an easier way to prove that $E^c\cap E = \varnothing$: you could try proving it by assuming that there is an element $x$ in $E^C\cap E$ and reaching a contradiction. Using this technique, you could have ...


3

Yes! For every ordinal $\alpha$, there is a limit of limits less than $\omega_1$ but greater than $\alpha$. So by regularity, there cannot be only countably many. This argument works for any unbounded set, actually. Proving that sets are unbounded can be done manually, or with the help of some theory of stationary and club sets.


3

Here is an idea which should work. Let $n >0$ be a positive integer. Define $k_0=n$, and recursively, as long as $k_j \neq 0$ define $$ k_{j} =2^{k_{j+1}} m_{j+1} \, \mbox { with } m_{j+1} \mbox{ odd } $$ This process ends after $t$ steps, when $k_t=0$, or equivalently $k_{t-1}$ is odd. Now define $$f(n) =( \frac{m_1-1}{2}, \frac{m_2-1}{2},.., ...


2

If something is not in $B \cup C$, then it is in neither $B$ nor $C$. Because if it was in $B$, then it is in $B$ or $C$, which is $B \cup C$. So the first mistake is in the second sentence. You could show that $A - (B \cup C) \subseteq A - (B \cap C)$. Because the former one is what is in $A$, but not in $B$ nor $C$. The latter is what is in $A$, but not ...


2

That's a very creative idea. I don't know of an established set theory symbol for this (e.g. http://www.rapidtables.com/math/symbols/Set_Symbols.htm doesn't even define such an "operation") and it doesn't seem to be a standard symbol in $\LaTeX$. One problem I can see is that the symbol $\ominus$ is quite widely used to denote the symmetric difference of ...


2

It's not awful, at least as an idea, but I would certainly say it's unnecessary, and I strongly dislike your particular notation choice. Basically it's a symbol that's too different and unintuitive to have to remember for too unimportant and uncommon a situation to be worthwhile. In my experience, once one gets to mathematics classes, textbooks, etc. that ...


2

In your post you only proved that $B = \overline A \cup X \implies A\cap B \subseteq X \subseteq B$, although you're missing the parenthesis in the expression $(\overline A \cap \overline X)\cup \overline A \cup X$. After parenthesizing it, your equality still holds: \begin{align} (\overline A \cap \overline X)\cup \overline A \cup X &= ...


2

Saying $A\subset B$ or $A\subseteq B$ means that $A$ is a subset of $B$. Some people use $A\subset B$ meaning that $A\subseteq B$ with $A\neq B$. Another notation for this is $A\subsetneq B$. Writing $B \supset A$, $B\supseteq A$ and $B\supsetneq A$ is equivalent to the above discussion. There is no difference between $\subseteq$ and $\subseteqq$, etc.


2

It means "is a superset of" as in $\{1,2\}\supseteq\{1\}$.


2

By "almost equal", you mean "almost everywhere equal". Which means that the two functions agree except possibly on a set of measure zero. You can kind of visualize the measure of a subset of the real line as being its "length". And in fact, this is exactly what the measure is for intervals: that is, the measure of $(a,b)$ is $b-a.$ Now, what we call ...


2

Here is a Venn diagram solution, masterfully drawn in Paint. Top: The three sets. Bottom left: $(A-B)-C$ in black. Bottom right: $A-(B-C)$ in black. We can easily see that the two differ by exactly $A\cap C$.


2

Context is everything. What is the set $\{x\mid x\neq 0\}$? If you are working in $\Bbb Q$ the set is different from the set you obtain if you work in $\Bbb R$ or in $\Bbb Z$ or in $\Bbb C$. If you are working inside "the mathematical universe", and you can "access" every mathematical object, then this is not a set anymore, since the mathematical universe ...


2

$B \cup C$ is all elements that are in either $B$ or $C$. So $A - (B \cup C)$ are all elements in $A$ but not in either $B$ or $C$, so in $A$ but in neither $B$ nor $C$. $A - (B \cap C)$ would be all elements in $A$ but not in $B$ and $C$. $(A - B) \cap C$ would be all elements that are in both $A$ and $C$ but not in $B$. $(A - B) \cup C$ would be all ...


2

Some examples include: The set $\mathbb Q$ of rational numbers with the usual ordering, which is interesting in that it has the same cardinality as your $A$ but a very different kind of order (a dense order). The long line, which is interesting in that it has the same cardinality as the set $\mathbb R$ of real numbers but a substantially different kind of ...


2

The order of the quantifiers is important. The statement as you have interpreted it would be written $$(\exists B\in P(X))(\forall A\in P(X))(271\in A\Delta B).$$ That is, there exists a fixed set $B$ such that for every set $A$, we have $271\in A\Delta B$. In other words, first you fix $B$, then you have to check it against all $A$. But the statement as ...


1

If 271 belongs to A ,let B be the empty set.If 271 does not belong to A, let B=X.


1

This is partly a question of quantifiers: You are trying to prove "For all A, there exists a B such that ...". From the definition of $A \Delta B$, you should be able to prove that $B = (A \Delta B) \Delta A$ So for any $A$, you are free to choose an $A \Delta B$ that contains $271$ (or any element, or elements, you choose), and the appropriate $B$ is ...


1

If I understand you correctly. You have $(F_i)_{i\in I } $ and $(G_i)_{i\in I}$ are families of set and you want to proof $$(\forall i \in I \ \ F_i \subset G_i ) \Longrightarrow \bigcup_{i\in I} F_i \subset \bigcup_{i\in I} G_i $$ Using proof by contradiction : suppose that $ \bigcup_{i\in I} F_i \not\subset \bigcup_{i\in I} G_i$ this implies that there ...



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