The interval between to numbers $[a,b]$ includes all the real numbers between and include $a$ and $b$. Is there a similar notation for a discrete interval, lets say for $\{a,a+1,a+2,\cdots,b-2, b-1, b\}$?
105625
The well-known notation for continuous intervals is $[a,b]$. But what's the case for discrete intervals? Actually they are sets of finite elements $\left\{a, a+1, ..., b-1, b\right\}$ or infinite elements $\left\{0, 1, 2, ...\right\}$.  Is there any special notation or common practice for discrete intervals?
23130
Prove that Real Numbers and the interval $(0,\infty)$ have the same cardinality.   Attempt:  Consider the function $f(x) = e^x$.   The domain of this function is all real numbers.  The range of this function is from $0$ to infinity.   Let $e^a = e^b$.   Then $\ln(e^a) = \ln(e^b)$  Then $a\ln(e) = b\ln(e)$  This means that $a = b$  Hence, $f$ is injective.   Let $c &gt; 0$   Then $e^{ln(c)} = c$  Since $c &gt; 0, \ln(c)$ is defined, so $f(\ln(c)) = c$  Therefore, f is surjective.   Then f is bijective.   Hence, $\mathbb{R}$ and $(0, \infty)$ have the same cardinality.
100051
I know $[0,1]$ denotes the interval between 0 and 1 with the boundary. And $(0,1)$ denotes the interval between 0 and 1 without the boundary. Today, I encounter some expressions as $]0,1]$, $]-\infty,+\infty]$. What does $]0,1]$ mean? Thanks.
2276
A) Specify a bijection from [0,1] to (0,1]. This shows that  |[0,1]|   =  |(0,1]|    B) The Cantor-Bernstein-Schroeder (CBS) theorem says that if there's an injection from A to B and an injection from B to A, then there's a bijection from A to B (ie, |A| = |B|). Use this to come to again show that   |[0;1]| = |(0;1]|
49101
Just like the title says, what is the notation for specifying the fractional part of a number?
23980
I'm reading .     The graph of $f: A \rightarrow B$ is the subset of $A × B$ consisting   of all ordered pairs $(a, b)$ with $f (a) = b$. If $A$ happens to be   $B^n$ for some $n ∈ N$, then we say that $f$ is a function on $B$ and   $n$ is the arity of $f$.   I don't understand the notation for the exponent of a set, I'm aware of what  is, it's the number of arguments or operands the function takes. I'm just not so sure if I can assume that a set with an exponent such as in $B^n$ means that a function has two operands.  I know the question may be too naive and perhapes the answer is the one I suggested and it's right in front of my face, I'm just searching for a confirmation or the negation of it. Thanks.
10350
Suppose we have graph $G  =(V, E)$ that $V(G) = {a, b, c, d, e}$ and $E(G) = {ab, ad, bc, be, cd, de} $  How should I know if this is an interval graph or not?
67644
So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n &gt; \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily).
5356
Use an induction argument to prove that for any natural number $n$, the interval $(n,n+1)$ does not contain any natural number.    I don't know where I could go with an induction argument. I was thinking of proving that if $s\in (n,n+1)$, where $s$ is a natural number, then $s-n$ is a natural number which lies in $(0,1)$, which is impossible as all natural numbers are bounded below by $1$.   However, this assumes that natural numbers are closed under addition. Also, this does not use induction.   Any pointers for this question?
16548
I have frequently encountered both $\langle a,b \rangle$ and $[a,b]$ as notation for closed . I have mostly encountered $(a,b)$ for open intervals, but I have also seen $]a,b[$. I recall someone calling the notation with $[a,b]$ and $]a,b[$ as French notation.   What are the origins of the two notations?  Is the name French notation correct? Are they used frequently in France? Or were they perhaps prevalent in French mathematical community at some point? (In this  Bourbaki is mentioned in connection with the notation $]a,b[$.)      Since several answerers have already mentioned that they have never seen $\langle a,b \rangle$ to be used for closed intervals, I have tried to look for some occurrences for this. The best I can come up with is the article on , where these notations are called Czech notation and French notation. Using $(a,b)$ and $[a,b]$ is called English notation in that article. (I am from Central Europe, too, so it is perhaps not that surprising that I have seen this notation in lectures.) I also tried to google for  or . Surprisingly, this lead me to a  where this notation is used for open interval.
7403
In my discrete mathematics class our notes say that between set $A$ (having $6$ elements) and set $B$ (having $8$ elements), there are $8^6$ distinct functions that can be formed, in other words: $|B|^{|A|}$ distinct functions. But no explanation is offered and I can't seem to figure out why this is true. Can anyone elaborate?
25990
Let $x,y,a,b$ be real numbers.  For $x&lt;y$ and $a&lt;b$, show the cardinality of $[a,b]$ equals the cardinality of $[x,y]$.  I did the above problem, by defining a linear function as a bijection between the intervals. The next part of the problem states,     Extend your result (from above) to open and half-open intervals.    I'm just not sure what exactly I'm supposed to do next, the question seems vague. Am I supposed to show $(a,b)\sim (x,y)$? $[a,b]\sim (x,y)$? $[a,b)\sim (x,y]$?
81405
Define a number $q$ in binary notation whose $n$-th bit is $1$ for $n$ prime, and $0$ for $n$ composite. So its 2nd, 3rd, 5th, 7th, 11th, etc. bits are $1$, with all other bits $0$. Here is $q$ out to its $101$-st bit: $$.01101010001010001010001000001    010000010001010001000001000    001010000010001010000010001    000001000000010001$$ What is known about $q$? ($\approx 0.414683_{10}$). Has it been investigated? Does it have a name? It is irrational. But is it transcendental?
57818
Wat exactly does the notation $$A : (BC)$$ mean? We are talking about fractional R ideals. From definitions i get, if $$a \in A:(BC)$$ then $$a(BC)\subset A$$  But what am i saying exactly?  I hope someone can explain this to me.
46231
What is the definition of an interval and why is the empty set an interval by that definition?
117621
a. Prove that the interval $A = [1,3]$ has the same cardinality as $B = [1,5]$ by writing down a bijection from $A \to B$. Don't prove it is a bijection.  b. Consider the following infinite set: $A = {1, \frac{1}{2},  \frac{1}{3},  \frac{1}{4},..., \frac{1}{n}}$,     Prove the set $A$ has the same cardinality as the integers by writing down a     .  bijection from $A$ onto $Z$.  I don't know how to find a function that is a bijection from one set to another. Can anyone help by explaining the thought process behind it, I've been having trouble with these types of problems? Thanks.
118302
I'm trying to prove that any proper interval has the same cardinality of the reals numbers $\mathbb R$. In order to prove this I define two functions $f:(s,t)\to (u,v),f(x)=\frac{v-u}{t-s}(x-s)+u$ and $g:\mathbb R\to (-1,1),g(x)=\frac{x}{1+|x|}$.  My question is are these functions bijections and from that can I conclude any interval has the same cardinality of $\mathbb R$?  Thanks in advance