Show that if $R$ is a partial ordering on a set $A$, then there exists a linear ordering $R^*$ on $A$ such that $R \subseteq R^*$
20551
How do I show that every partial order can be extended to a linear ordering?      I think that I manage to prove that claim for finite set, how can I prove it for infinite set?  Thank you.
105159
Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.   proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.  I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated.
20035
Let u and v be elements of a commutative group, and suppose that their   orders are r and s, respectively. Show that if r and s are distinct   primes then the order of uv is rs.   I will use the follwing theorem to proof this.  Theorem: Let x be an element of order m in a finite group G. Then $$x^s=1$$ in G if and only if s i a multiple of m.    My attempt  We can here see that $$(uv)^{rs}=[commutative]=u^{rs}v^{rs}=1^s1^r=1$$ By the theorem, this tells us that this is true only if only rs is a multiple of the order of $uv$. Let the order be $ord(uv)=k$. We know that every positive integer can be written as $$a=bm+l, 0&lt;=l&lt;b$$ and since $r$ and $s$ is primes (positive integers), $rs$ can be written as $$rs=km+l, 0&lt;=l&lt;k$$ This gives ut that $$1=(uv)^{rs}=(uv)^{km+l}=1^muv^l=(uv)^l$$ If l>0 then the equation $(uv)^l=1$ contradicts that the positive integer k is the least positive integer for which $(uv)^k=1$. Therefore l=0 and $rs=km$ . We can here see that $rs$ is a multiple of k (and this proof also the theorem).  $$rs=km =&gt; k | rs =&gt; k|r     ORk|s$$ But we know that $r$ and $s$ are distinct primes, so k must be equal to one of the primes.  $$k=r =&gt; (uv)^k=(uv)^r=1v^r=v^r=1$$ And since we know that $ord(v)=s$ the theorem tells us that $r$ must be a multiple of $s$. But since both of them is primes, we get that $r=s$ but this is a contradiction (r and s are distincts as the exercise tells us).   Now, we try with $k=s$. Same as above, we get that $s=r$ which is a contradiction. Neither k=r nor k=s is right since by the theorem we get that $k=r=s$ and we have that $r$ and $s$ are distincts. This tells us that k cannot be equal to r nor s, since this gives us that $r=s$. And since we know (and as i did proof)that  $$rs=km$$, this tells us, as i wrote before, that $$k|rs$$ Since k is neither equal to $r$ or $s$, this tells us that k must be equal to $rs$. Therefore $$ord(uv)=k=rs$$.  Edit: i know that this can be prooved by l.c.m(x,y) but we have not used it so far. That's why i did not proof it with l.c.m(x,y)
121107
Let $(X_{\alpha})_{\alpha \in A}, (Y_{\alpha})_{\alpha \in A}$ be families of sets such that for $\{X_{\alpha} \mid \alpha \in A \}$ and $\{Y_{\alpha} \mid \alpha \in A \}$ are totally ordered by inclusion. We have $(X_{\alpha},Y_{\alpha}) \neq (X_{\beta},Y_{\beta})$ for $\alpha \neq \beta$. Suppose that $|X_{\alpha}| \leq |Y_{\alpha}|$ for any $\alpha \in A$. It is true that $|\bigcup_{\alpha} X_{\alpha}| \leq |\bigcup_{\alpha} Y_{\alpha}|$?
24139
Let $f:A \to \mathbb R$ and suppose that $$ \sup\Big\{\sum_{a\in F}\lvert\, f(a)\rvert :  F\text{ is finite subset of }A\Big\} &lt; \infty $$ then the set $\{ a \in A : f(a) &gt; 0\}$ is countable.  My try: Define these element of ${ a ∈ A : f(a) &gt; 0}$, then $∑_{n=1}^\infty\lvert\, f(a_n)\rvert$ is determined, regardless of the method of arrangement.
154
I need to prove the following letting A and B be subsets of a universal set U. $$A\cup B = U \iff A^c\subset B$$ It just isn't straight forward to me.
21292
Consider this statement: "every set can be linearly ordered." Can we prove it without AC?
5674
Let's say I take a set $S$, where $S$ can be well ordered. From what I understand, one can use that well ordering to totally order $\mathscr{P}(S)$.  How does a body actually use the well ordering of $S$ to construct a total ordering of $\mathscr{P}(S)$?  Perhaps construct is too strong a word. Is there a way to prove that a total ordering of $\mathscr{P}(S)$ just exists, no need to explicitly state what it is, since that might be very difficult.
24061
It's true that a set with a linear order where every initial segment is finite is necessarily countable? I don't think so but I can't find a counterexample (or make a proof)
51320
Let $(X,⪯$) be totally ordered set, prove that if every non empty countable subset of $X$ is well ordered then X is well ordered.   It does seem obvious that any subset should have a minimum but I am not sure of how to prove it. Any help would be appreciated, and sorry I put up the same problem earlier but forgot to include the 'countable' subset.
16848
I was thinking if there is a solution to this problem without trying to explicitly create Vitali sets in $A$. Does anyone have any ideas?
46211
Let $R$ be a total order on set $S$. Prove that if $S$ has a minimal element, than the minimum element is unique.  I have difficulties with proofs. I know any graph of a total order is a straight line, which clearly has a minimal element. How doI tell when something is mathematically proven?  EDIT: this is how I would prove it  Let $x, y \in S$ such that x and y are minimal. Since S is a total order $x$ and $y$ are comparable and either    case 1: one comes after the other and therefore one isn't the minimal case 2: both are the same and $x=y$  I know this isn't a mathy argument and could use help making it more presentable.
107195
I need a hint for the following question:     Let $S$ be a nonempty ordered set such that every nonempty subset $E\subseteq S$  has both a least upper bound and a greatest lower bound. Suppose $f:S \rightarrow S$ is a monotonically increasing function. Show that there exists an $x\in S$ so that $f(x)=x$.   My reasoning so far has been as follows:   I wanted to prove this by contradiction. If we assume the statement is false, this implies that for each $x\in S$, $f(x) &gt; x$ or $f(x) &lt; x$. Let $A:=\{x:f(x)&lt;x\}$, $B:=\{x:f(x)&gt;x\}$.  Let's assume $A$ is empty. Then it is easy to show a contradiction. We know that $\sup(B)\in B$. But that means that $f(\sup(B))&gt;\sup(B)$ which is impossible. The same idea applies if we assume $B$ is empty.   What I am trying to do is prove the case where both $A$ and $B$ are nonempty. To prove the first part I haven't even used the monoticity of the function $f$, so I know that I have to use it at this point. That means trying to find a contradiction by finding $x,y\in S$, $x\le y$, $f(x)&gt;f(y)$. I was trying to look at the supremum and infimum of $A$ and $B$ but that didn't seem to lead me anywhere.  I would appreciate hints!
89471
Suppose $x\in\mathbb{R}^n$ and $A\in\mathbb{R}^{n\times n}$. What is a sufficient condition for $$x, Ax, A^2x, ..., A^{n-1}x$$ being linearly independent?
83799
Can someone please verify this?     Show that if $B \subseteq C$, then $\mathcal{P}(B) \subseteq \mathcal{P}(C)$   let $x \in \mathcal{P}(B)$.  Then, $x \subseteq B$  This implies that $$\forall a \in x, a \in B$$  This further implies that $$\forall a \in x, a \in C$$  Therefore, $$x \subseteq C$$  This implies that $$x \in \mathcal{P}(C)$$
92238
Let $n \in \mathbb N$ and $U\subset \mathbb R^n$ be an open set.Let $K\subset U$ be a compact subset of $U$. Then How can we show that distance between $K$ and $\partial U$ is positive?   Please, someone, give some hints. to solve this.  Thank you.
48529
Could anyone provide an example of a nonempty subset $U$ of $R^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $R^2$