Let $A$ and $B$ be two subsets of a vector space $V$ and assume that $A \subseteq B$. Prove that $Span(A) \le Span(B)$.  What I tried: to prove that $Span(A)$ is a subset of $Span(B)$, I tried to use the definition of spanning sets. I wasn't sure how to prove that + and * are closed for $Span(B)$.
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If X and Y are two sets of vectors in a vector space V, and if X $\subset$ Y, then is span X $\subset$ span Y? If so, why is or isn't the span of X a subset of the span of Y?  EDIT:  Thank you for the hints! Here is the proof I came up with; please let me know if it is correct.    The spanning set of X can be written as: Span(X) = {$a_1 x_1 + a_2 x_2 + ... + a_n x_n$} where all $x_i$ are vectors and all $a_i$ are scalars. Since X$\subset$Y then all $x_i$ are also in Y. Thus, Span(x) is a linear combination of vectors from Y So Span(x) $\subset$ Span(Y).
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 $$x \in \mathcal{P}(C)$$
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How would one denote the set of all subsets of $A$ which have size $2$?  Would  $$\binom{A}{2}$$  be a good idea?
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I am trying to define a proof for  $A \cup B = B$ if and only if $A \subset B$  I started with:  $$A \cup B$$ $$x \in A \lor x \in B$$   but I am unsure on how to continue.
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I was doing some linear algebra exercies. One I did is this: Prove that the union of two subspaces $U,W$ of $V$ is a subspace of $V$ if and only if one contains the other.   My proof is this: If one is a subset of the other $U \subset W$ then the union is $W$ which by assumption is a subspace. If neither is contained in the other there exists an example where the union is not a subspace: let $V$ be the xy-plane, $U$ the x-axis and $W$ the y-axis. Then the union is not closed with respect to addition.   My question is whether it is valid to prove something by example? How would you prove this in another way?
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I came to know about 2 definitions of a subbasis of a topology. 1. A collection of subsets of X viz. S is said to be a subbasis of a topology T on X if S covers X and the finite intersections of elements of the subbasis form the basis elements (H.L.Royden). 2. If we consider the all the topologies on X containing S, then their intersection is also a topology on X and it is the smallest topology containing S. Then S is said to be a subbasis of the smallest topology, and the topology is said to be generated by S.  Now I want to prove their equivalency, i.e. S is a subbasis of a topology T iff T is generated by S. I have tried in the way: First let T is generated by S. Then by 2nd definition T is a topology containing S. Let B be the set of all finite intersections of elements of S.Then S is a subset of B and B is a subset of T (as T is a topology). LET T1 be the collection of possible unions of elements of B, we will show that T = T1. After this I am not able to proceed. Also I am unable to sketch the converse part. If something is wrong, please let me know, and also I hope to get help. Thank You.
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Consider the following putative theorem.   Theorem? Suppose A, B, and C are sets and $A \subseteq B \cup C$. Then either $A \subseteq B$ or $A \subseteq C$.   What's wrong with the following proof?   Proof. Let $x$ be an arbitrary element of A. Since $A \subseteq B \cup C$, it follows that either $x \in B$ or $x \in C$.   Case 1. $x \in B$. Since x was an arbitrary element of A, it follows that $\forall x \in A(x \in B)$, which means that $A \subseteq B$.  Case 2. $x \in C$. Similarly, since x was an arbitrary element of A, we can conclude that $A \subseteq C$.  Thus, either $A \subseteq B$ or $A \subseteq C$.   I have proved that the theorem is incorrect but I can't understand why the above proof is not correct.
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What does it mean when you have a set like A = {1, {2}, 3, 4}? Is {2} a subset within A?  Say I have the following:  A = {1, {2}, 3, 4}, B = {1, 2, {3}}, C = {1, 2}  How many elements are in $A \cap B \cap C$? 1 or 2?  What if I want to find $A \times B$, is there 6 elements or 12?
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Let $A$ be an $n\times n$ matrix. Prove that $\dim(\mathrm{span}({I_n,A,A^2,...})) ≤ n$  I'm at a total loss here... Can someone help me get started?
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Task is:  Let $A$ be a finite set and let $B$ be a subset of $A$ with $|A|=n$,$|B|=m$ and $0&lt; m &lt; n$.Find a formula for the number of subsets of $A$ that contain $B$ and prove your statement.  I assume this is $2^m$ + something. $2^m$ would count the subsets containing only elements of $B$, but i'm not sure how the sets of the form $(a_1,a_2,b_1)$ or similar would be counted. I'm having troubles with this so help would be appreciated.
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A question from my homework I'm having trouble understanding. We are given:  $A(1) = \{\varnothing\}$, $A(n+1) = A(n)\cup (A(n)\times A(n))$  $A=A(1)\cup A(2)\cup A(3)\cup \cdots \cup A(n)\cup A(n+1) \cup \cdots$ to infinity  The questions are:  1) show that $A\times A \subseteq A$  2) Is $A \times A = A$?  Thank you for your help.  I've tried writing $A(2)$ but it gets really complicated and I'm having trouble understanding what the sets are. Let alone solve the question.
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This is what I have so far:   First, we will attempt to show $A-(B-C) \subseteq (A-C) - C.$     Let $x \in A -(B-C)$.      Then $x\in A$ and $x \notin (B-C)$. By DeMorgan's we have that  $$x \notin (B-C) = x \notin B \vee x \in C.$$     We have that $x \in A \wedge (x \notin B \vee x \in C)$.      Then we have that $(x \in A \wedge x \notin B) \vee (x \in A \wedge x \in C)$.  So by definition of $\cup$ and $\cap$, we have $x \in (A-B) \cup (x\in (A \cap C))$. I'm kind of stuck here.  I can see a contradiction here in that LHS says $x \in C$ but the RHS says $x \notin C$.   Can anyone tell me what to do from here? I'm not quite sure how to formally state that the statement is false. I feel like I'm not being complete here.
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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.
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An exercise asks me to prove the following: $$A \cap (B-A) = \varnothing$$ This is what I did:  I need to prove that $A \cap (B-A) \subseteq \varnothing$ and $\varnothing \subseteq A \cap (B-A)$  The second one seems to be obvious by definition (not sure if it is fine to say that in a test).  But the first one goes like this: $$x \in A \cap (B-A)$$ $$x \in A \land x \in (B-A)$$ $$x \in A \land (x \in B \land x \notin A)$$ $$(x \in A \land x \notin A) \land x \in B$$ $$(x \in A \land x \notin A)$$ Since there is a contradiction... err... the proof... is good.  Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of $A \cap (B-A) \subseteq \varnothing$
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Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$?
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Prove that any two bases of a finite dimensional vector space must have the same number of elements.  By considering the following two bases  $$S_1 = \{ \alpha_1, \alpha_2 , \ldots, \alpha_n \},$$  $$S_2 = \{ \beta_1, \beta_2, \ldots, \beta_m \},$$  how do I show that $m = n$?  Hints to get started. Thanks very much.
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I need help with this proof:     Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.   My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.  Where do I go from here?