Let $A$ and $B$ be two closed subsets of the set of real numbers.  Define $A+B=\{a+b\in\mathbb{R}:a\in A ,b\in B\}$.  Is it true that $A+B$ is closed in $\mathbb{R}$?  If not, could you give a counter-example?
1030
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?  (By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)  Thanks!
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 $(a,b)\sim (x,y)$? $[a,b]\sim (x,y)$? $[a,b)\sim (x,y]$?
11662
In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?  I tried by assuming a set is equal to its interior points and contains its limit points.  A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?  Anyway, it must then be unbounded.  If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$.
22499
I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$. Can you provide me some hints please?
86878
$Q$: Find a subset of $ \mathbb R^3$ which is closed under $*$, but not $+$. $A$: The set $\mathbb R$$u$ $+$ $\mathbb R$$v$, where $u$ and $v$ are linearly independent vectors in $ \mathbb R^3$. Here $ \mathbb R$$w$ = $\{aw: a \in \mathbb R\}$ for a given vector $w$. This is closed under $*$, but not $+$.  By  $u$ and $v$ are linearly independent vectors in $ \mathbb R^3$  do they mean two sets of linearly independent vectors $u$ and $v$ in $ \mathbb R^3$ where $u$ could be  $u = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$? Generally, please help me understand how this set is not closed under $+$. Thanks.
121294
Let $A\subset \mathbb{R}$ be of positive Lebesgue measure, i.e. $\mu(A)&gt;0$. Is it then true that $\mu(\mathbb{R}\setminus (A+\mathbb{Q})) = 0$?   I am quite sure that if $\mu(A)&gt;0$, then $A-A$ contains a rational, as well as an irrational number, but I'm not sure if this actually helps.
28012
I am working on this exercise for an introductory Real Analysis course:     Show that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   I know that $\mathbb{R}$ is uncountable. I also know that two sets $A$ and $B$ have the same cardinality if there is a bijection from $A$ onto $B$. So if I show that there exists a bijection from $\mathbb{R}$ onto $\mathbb{R}^2$ then I beleive that shows that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   Let $x_i \in \mathbb{R}$, where each $x_i$ is expressed as an infinite decimal, written as $x_i = x_{i0}.x_{i1}x_{i2}x_{i3}...,$. Each $x_{i0}$ is an integer, and $x_{ik} \in \left \{ 0,1,2, 3, 4, 5, 6, 7, 8, 9 \right \}$. Then, let  $$f(x_i)=(x_{i0}.x_{i1}x_{i3}x_{i5}... ,x_{i0}.x_{i2}x_{i4}x_{i6}...)$$  What should I do to show that $f: \mathbb{R} \to \mathbb{R}^2$ is an injective function? Any suggestions or help with the question would be appreciated.
92059
$\Bbb Z$ - integers $\Bbb N$ - natural numbers (starting from 1) $\Bbb R$ - real numbers    I believe the answer is the set of real numbers ($\Bbb R$), seeing as $b$ will not equal $0$ as the set of natural numbers start from $1$. Thoughts?
73289
Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.
42854
Please let me know if you find my proof satisfactory.  Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.  If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$. If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon &gt; 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.
86031
Is it correct to say that $\mathbb{R}$ is a subset of $\mathbb{R}^2$? Or, put more generally, given $n,m\in\mathbb{N}$, $n&lt;m$, is $\mathbb{R}^n$ a subset of $\mathbb{R}^m$?  Also, strictly related to that: what is then the "relationship" between the set $\{(x,0)\in\mathbb{R}^2,x\in\mathbb{R}\}\subset\mathbb{R}^2$ and $\mathbb{R}$? Do they coincide (I would say no)? As vector spaces, do they have the same dimension (I would say yes)?  If you could give me a reference book for this kind of stuff, I would really appreciate it.  Thank you very much in advance.  (Please correct the tags if they are not appropriate)
49868
In $\mathbb{R}$ with standard topology the closure of the above set $M$ is $M \cup 0$. I fail showing this by purely topological means. I.e. not using the fact that this holds in the reals as a metric space and deduce it for the induced topology.  I assumed that there exist a closed subset $A$ containing $M$ but not 0. Hence, the complement of $A$ is an open subset containing 0. Does this already yield a contradiction or do I need a different approach?  Could you help me on this?
110867
Is the set of integers $ \mathbb{Z}$ closed in $\mathbb{R}$ equipped with the usual topology?
66674
I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion.  The set of real numbers $\mathbb{R}$ is obtained as completion of $\mathbb{Q}$. However, $\mathbb{Q}$ is not the only set which is dense in $\mathbb{R}$. $\mathbb{Q} \backslash \mathbb{Z}$ is also a dense subset of $\mathbb{R}$. I am wondering if it makes sense to talk of the "smallest" dense subset of $\mathbb{R}$. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set $A$ of $\mathbb{R}$ such that if $B$ is a proper subset of $A$ then $B$ is not dense in $\mathbb{R}$. I am able to "see" that the set $A$, I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why $A$ doesn't exist. Could some one throw more light on this?
104753
If there is a closed subset $B$ of $\mathbb{R}^d$ and $z\in \mathbb{R}^d$ then show there is a point $b_0\in B$ so that we have $\delta(z,B) = \|z-b_0\|.$  I'm not sure how to begin this exercise and what theorems are available to help me solve it
122027
"Ring" and "semiring" are concepts defined both in algebra and set theory.  In Algebra  A  is a set R equipped with two binary operations + and · called addition and multiplication, that Addition (+) is abelian, Multiplication (⋅) is associative, Multiplication distributes over addition, and Multiplicative identity (1) exists.  A ,  is an algebraic structure similar to a ring, but without the requirement that each element must have an additive inverse.   In Set theory  A  is a family of sets closed under unions and set-theoretic differences.  That is, it obeys the two properties $$A \setminus B \in \mathcal{R} $$ $$A \cup B \in \mathcal{R}$$ This implies that it is also closed under intersections, $$A \cap B \in \mathcal{R}$$  A  is a non-empty collection S of sets such that   $\emptyset \in S$ If $E \in S$ and $F \in S$ then $E \cap F \in S$. If $E \in S$ and $F \in S$ then there exists a finite number of mutually disjoint sets $C_i \in S$ for $i=1,\ldots,n$ such that $E    \setminus F = \bigcup_{i=1}^n C_i$.   I wonder, are such definitions somehow linked or equivalent? or they are just coincidence that both fields of mathematics used the same terms?
96502
Given two sets $S_1$ and $S_2$ in $\mathbb R^n$ define their sum by    $$S_1+S_2=\{x\in\mathbb R^n; x=x_1+x_2, x_1\in S_1, x_2\in S_2\}.$$   Prove that if $S_1$ and $S_2$ are compact, $S_1+S_2$ is also compact.   Prove that the sum of two compact sets in $\mathbb R^n$ is compact.   Compact set is the one which is both bounded and closed. The finite union of closed sets is closed. But union is not the same as defined in the task. I so not know how to proceed. I do understand that I need to show that the resulting set is both bounded and closed, but I do know how to do that.