Hot answers tagged

13

You are conflating end points and points. The circumference of a circle does not have any end points but infinitely many points. The line segment has two end points and still infinitely many points. Sidenote: The length of the segment is not as important as you may think. We cannot distinguish the number of points on a line segment of length $π$ from ...


9

You can prove this using induction. Our trivial base case is $n=2$. Now, assume that the statement is true for $n-1$ We can split a $2^n\times 2^n$ board into four $2^{n-1} \times 2^{n-1}$ boards, as shown above. W.L.O.G. the square that we choose to take out is in the upper right quadrant. Since a $2^{n-1} \times 2^{n-1}$ board can be filled with Ls ...


8

Your set is actually equal to $(2, \infty)$, since $(0,0)=(1,1)=\varnothing$, but even the set $\{ 0, 1 \} \cup (2, \infty)$ is not equal to $\mathbb{R}^{\ge 0}$: for example, $\frac{1}{2}$ and $\frac{3}{2}$ are not elements of this set.


7

The nifty thing about sets is that they are things. We can talk about functions whose values are sets, or functions that take sets as inputs, or make sets of sets, or sets of sets of sets ... This is quite useful, and in fact absolutely ubiquitous in higher mathematics. The downside of this is that when sets are "things", there can't be a set for any ...


6

$\varnothing = \{\}$ It is the empty set.   It is not just nothing.   It is a set with no elements. Sets can be elements of other sets.   The set of the emepty set is not empty, it contains the empty set.   $\{\varnothing\}=\{\{\}\}$ So we have that: $\{1,2\}\cup\{\} = \{1,2\}$ but $\{1,2\}\cup\{{\small\{\}}\} = \{1,2,{\small\{\}}\}$ ...


6

It is obviously true for $2\times 2$. As an advice, when you see $2^n$ try to divide it into two parts consisting of $2^{n-1}$ and use induction. In your case, divide the whole board into $4$ $2^{n-1}\times 2^{n-1}$ boards. Edit: Without loss of generality, suppose we remove a square in the up-right quadrant. Then, on the up-right we have $2^{n-1}\times ...


5

What you have defined is in fact the $\beth$ function. It is a continuous function, namely, it is increasing and the limit is evaluated as the limit of previous results. Therefore there is a proper class of fixed points. The fact that cardinal exponentiation returns a cardinal, and the supremum of cardinals is a cardinal ensures that such fixed points will ...


5

There are lots of different things in mathematics called "infinity" or "infinite". Infinite cardinals are different from infinite ordinals, and the $+\infty$ encountered in calculus is a different thing from those, and the $\infty$ that is neither $+\infty$ nor $-\infty$ but is approached by going in either direction on the line is slightly different from ...


4

Both the circumference and the segment have infinitely many points, but that's not the point (heh, heh). What matters is their length. The picture presumably is trying to show the ratio between the diameter and the circumference, which turns up to be $\pi$, or 3 and then something.


4

When in doubt, try to work out smaller examples. If $S$ is the union of $\{\{1\}\}$ and $\{\{2\}\}$. Which one of the three is correct? $S=\{1,2\}$, $S=\{\{1,2\}\}$, $S=\{\{1\},\{2\}\}$. Not sure? Let's try an even simpler example. $S$ is the union of $\{A\}$ and $\{B\}$, what is $S$? If you answered $\{A,B\}$, you're correct. Let's kick it up a notch. ...


4

I think you'd need the Generalised Continuum Hypothesis to prove that $\aleph_\omega$ is a fixed point for $F$, though I'm willing to be corrected by people more knowledgeable than me. To show that all such fixed points are cardinals, suppose $\gamma$ is a fixed point and not a cardinal. As you correctly point out, $\gamma$ must be a limit ordinal, so we ...


4

Cardinals measure the number of elements in a set. Since there is no such thing as half an element of a set, there is absolutely no meaning to a reciprocal of a cardinal number. Certainly not an infinite cardinal. Similarly, ordinal numbers measure the order type of an ordered set. Specifically, a well-ordered set. Since in this case there is no meaning to ...


4

Your construction gives $[-1,1] \times [-1,1]$ because the two sets you've written down are those two closed intervals (and the set product is every pair of things where the first is from the first interval and the second is from the second interval). However, there is a stupid way to do this ... Let $A$ be a circle of radius $1$ and $B = \{0\}$. Then $A ...


3

So you have proved that "$A \subseteq B$ and $x \in A \cup B$ implies $x \in B$". You still need to prove that "$A \subseteq B$ and $x \in B$ implies $x \in A \cup B$" (which will complete the first direction of the equivalence), and then "$A \cup B = B$ implies $A \subseteq B$" (which is the other direction). To reiterate your first direction: If $A ...


3

You’re on the right track. Since $|K|=\kappa$, you might as well replace $K$ by $\kappa$ and index the sets by ordinals less than $\kappa$. For each $\xi\in\kappa$ you have an injection $f_\xi:A_\xi\to\lambda$. Define $$f:\bigcup_{\xi\in\kappa}A_\xi\to\kappa\times\lambda$$ as follows: for each $x\in\bigcup_{\xi\in\kappa}A_\xi$ let ...


3

Yes, that looks right. For the single remaining step, once you know that $\{x\}\in\mathcal P(B)$, this is the same as $\{x\}\subseteq B$, which again means that $x\in B$, as needed.


3

Well, the number of subsets of a set is the number of combinations that can be made out of it's elements. Let's imagine that a set is like a pair of two tuples (or lists) of equal length. One list contains all of the items that are possible within a set (I.E. the universe). In our case that list is the set of 1001 items. Now, the second set is a tuple of ...


3

Yes, you are right. Generally, we have $2^\kappa > \kappa$ for any cardinal number $\kappa$.


3

No, it's like the empty product or the empty sum. $$\prod_{x\in\emptyset}x=1$$ $$\sum_{x\in\emptyset}x=0$$ $$\bigcap_{x\in\emptyset}x=G$$ and through similar reasoning one would get $$\bigcup_{x\in\emptyset}x=\emptyset$$ as the operation on the empty set is always the neutral element of that operation. The reason is such that you can say "any collection" ...


3

$$A\cup B=((A \triangle B) \cap A)\triangle B$$


3

Note that if $A\cap B=\varnothing$, then $A\cup B=A\mathbin\triangle B$. So if you can show that $A\setminus B\in S$, then $A\cup B=(A\setminus B)\mathbin\triangle B$. Next, observe that $A\cap(A\mathbin\triangle B)=A\cap((A\setminus B)\cup(B\setminus A))=A\setminus B$.


3

This is a consequence of a very useful lemma whose general proof is almost as simple as providing a fixed point in this special case. Let's say that a function $F \colon \operatorname{On} \to \operatorname{On}$ is normal iff it is strictly increasing (i.e. $\alpha < \beta$ implies $F(\alpha) < F(\beta)$) and continuous (i.e. $F(\lambda) = \sup_{\alpha ...


3

NO: the statement "there exists the set $S$ of all sets that are not elements of themselves" has been proved to be false. The "logical analysis" of Russell's Paradox starts from the seemingly harmless principle [called: unrestricted Comprehension Axiom] that: for any formula $φ(x)$ containing $x$ as a free variable, there will exist the set $\{ x ...


3

You are mistaken. $\{x \in \emptyset \mid p(x) \}$ is still a set; it's just empty. It has no elements, because as you point out, if it had an element then that element would lie in the empty set (a contradiction).


2

It is $$(A\cap B)\cup(B^c\cap A) = (B\cup B^c) \cap A= A.$$ Note that this result always holds whether $A$ and $B$ are disjoint.


2

Yes, there is a mistake. it should be $(-1,0)$ on the right. Or perhaps the left hand side is $(-1+1/n,1)$ rather than $(-1+1/n,0)$.


2

His answer is wrong in that he stated "let $\epsilon $ be the radius of an interval centered at c" implying such an interval exists. Then he claims "such that" giving conditions assuming what is to be proven. BUT It could have worked as this: "Intervals centered a $c $ of every possible radii exist (by simply taking the interval $(c-r,c+r)$ where $r $ is ...


2

Take $f$ the identity function. Let $m=2$ with $B_1 = [0,1]$ and $[1,2]$. Let $n = 3$ with $A_1= [0,1], A_2 = [0,1]$ and $A_3 = [1,2]$. Now your inequality reads $$\frac{1}{3} \geq \frac{1}{2}$$ and we have a contradiction. EDIT: This is a counterexample to the original question. For a counterexample to the modified one see the comments.


2

Your attempt is okay. Alternative for converse: $A\in\wp(A)\subseteq\wp(B)\implies A\in\wp(B)$, i.e. $A\subseteq B$


2

You have the ($\implies$) direction right. As for the converse, you're almost there: ( $\Longleftarrow$ ) If $x\in A$, then $\{x\} \subseteq A$, so $\{x\} \in \mathcal{P}(A) \subseteq \mathcal{P}(B)$; hence $\{x\} \in \mathcal{P}(B)$. This means the same thing as $\{x\} \subseteq B$; in turn, the last statement is equivalent to $x\in B$. Just expand the ...



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