I'm currently trying to write up a solution to the following problem:  If $ \displaystyle A_1 \supseteq A_2 \supseteq A_3 ... $ where each $ \displaystyle A_j $ is a non-empty, finite set of real numbers, prove (or disprove) the intersection $\displaystyle \bigcap_{n \in \mathbb{N}} A_n $ is non-empty and finite.  My thoughts: this is obviously true, as for some point onwards we must, for some $ \displaystyle N \in \mathbb{N}$, have the following equality for all $ \displaystyle k \ge N$: $\displaystyle A_k = A_N$.  If this did not hold, then the only way it could fail would be for us to have strict super-set relations (at some stage), which can't continue indefinitely, for we know $ \displaystyle A_1$ is finite, and hence each set is finite.  However, I'm having trouble writing this up as a proper mathematical rigorous direct proof; my thought for a proof is something like follows.  Proof. We prove this by contrapostion. It follows that we should consider either (i) the intersection is infinite, or (ii) the intersection is empty.   If (i), then as the intersection is a subset of each $\displaystyle A_j$ for every $ \displaystyle j \in \mathbb{N}$, it follows that, in particular, $ \displaystyle A_1$ must be infinte, because it has an infinite subset.   We now suppose (ii); then for any element, $ \displaystyle x $, we cannot have $ \displaystyle x \in \bigcap_{N \in \mathbb{N}} A_n$. In particular, if $ \displaystyle A_1 = \{a_1, ..., a_k\}$, we can find a $ \displaystyle j_1 \in \mathbb{N}$ such that $\displaystyle a_1 \not\in A_{j_1}$, and generally find a $ \displaystyle j_s \in \mathbb{N}$, with $\displaystyle j_s &gt; j_t$ if $ \displaystyle t &lt; s$ such that $\displaystyle a_s \not\in A_{j_s}$, where $ \displaystyle 1 \le s \le k$. In particular, for $ \displaystyle s = k$, by the fact the $ \displaystyle j's$ are increasing, it follows $ \displaystyle a_1, ..., a_k \not\in A_{j_k}$, and from the original superset relation, this set is empty. $ \displaystyle \square$  My concerns: I think this is a valid proof, but I'd much rather prove it directly by using the first observation that eventually the subsets of $ \displaystyle A_1$ are all equal. However, I'm finding this hard to explicitly, and basically end up writing the above argument to show that eventually the sets are empty.   How can one prove this directly?
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Statement of the problem: Prove that if $A_1 \supseteq A_2 \supseteq A_3 \cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n=1}^{\infty} A_n$ is also finite and nonempty.  My solution: $\bigcap_{n=1}^{\infty} A_n = \emptyset.$ Then at least one $A_n$ is disjoint from the rest - contradiction, since each $A_n \subseteq A_{n-1}.$ Next assume that the intersection has infinite cardinality - contradiction, since $A_n \supseteq \bigcap_{n=1}^{\infty} A_n,$ but we have $A_n$ finite.  This is a homework problem and this is what I will be turning in regardless of any answers I get. I would like to know if there is a direct proof of this claim. I had considered trying to show that $\exists x \in A_n$ for each choice of $n,$ which would show that the intersection is nonempty, but: a) I don't know how to do this, and b) I still would not be able to prove the intersection was finite except by contradiction.  Thanks!  edit: Moreover, we were told not to use induction on this problem. But if $A_1$ is finite, can't we reduce the collection of $A_n&#39;s$ to a subset of the power set of $A_1$ and then induct?
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This is a homework problem for class credit and I would appreciate a hint.  Let $A$ be an infinite set and suppose $x, y$ and $z$ are points in $A$. Show that $A$ has the same cardinality as $A\setminus\{x,y,z\}$.  So far I am trying to solve this for the more general case, let $A$ be an infinite set and $B$ be a finite subset of $A$, show that $A$ has the same cardinality as $A\setminus B$.  If $B$ is finite and $A$ is infinite, then $A\setminus B$ is infinite because if $A\setminus B$ were finite then $A = A\setminus B \cup B$, the union of two finite sets and hence $A$ would be finite.  Using Cantor-Bernstein we know that if there is a one-to-one function from $A$ to $A\setminus B$ and a one-to-one function from $A\setminus B$ to $A$, then $A$ and $A\setminus B$ are equivalent.  It is easy to define a function $f:A\setminus B \to A$ that is one-to-one, $f(x) = x$.  Now I need to define a function $g:A \to A\setminus B$ that is one-to-one.  I have a feeling that this involves using the axiom of choice.  Is this the right strategy?  Have I made any errors?  And can anyone give me a hint on how to define a one to one function from $A$ to $A\setminus B$?
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what i had done :let X be a infinite set and there be open sets $A,B$ on it. such that $A\cap B =\varnothing$ or $A\cap B\ne\varnothing$then  $X-A\cap X-B \ne  \varnothing $ .which means there doesn't exist   $X-A\cap X-B = \varnothing $ ,which implies that there is no separation of X in finite complement topology. is this proof correct.
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This question is related to Exercise 2.15 in Baby Rudin, in which the reader is instructed to "[s]how that Theorem 2.36 and its Corollary become false (in $R^1$, for example) if the word 'compact' is replaced by 'closed' or by 'bounded.'"  The relevant theorem and corollary are as follows:  Theorem 2.36: If $\{K_{α}\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{α}\}$ is nonempty, then   $\{K_{α}\}$ is nonempty.  Corollary: If $\{K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n}$ contains $K_{n+1},  (n = 1, 2, 3, ...)$, then   $K_n$ is not empty.  My Question: I realize this question and similar ones have been asked elsewhere on MSE, but I've yet to see a straightforward (to me, at least) answer to my question specifically. Usually, for the "closed" part of the exercise, the following collection of closed sets, $\{K_n \}_{n \in \mathbb{N}}$ is used:  $$K_n = [n, \infty), n \in \mathbb{N}.$$  For, the intersection of every finite subcollection of ${K_n}$ is clearly nonempty. Indeed,  $$ \bigcap_{i=1}^{m} K_{n_{i}} = [N, \infty) $$  where $N = \max\{n_1, n_2, ..., n_m\}$.  Then, the assertion is made that   $$ \bigcap_{n=1}^{\infty} K_n = \emptyset. $$  However, I don't see how the same argument can't simply be used, i.e., that, for example, $N \in  \bigcap_{n=1}^{\infty} K_n$ (or maybe, say, $n+1$). Intuitively, this certainly makes sense, I'd say. For, given the archimedean property, it seems that as $n$ approaches infinity, we can always find an $l$ in each increasingly smaller set such that $n &lt; l &lt; \infty$; and so $l \in  \bigcap_{n=1}^{\infty} K_n$.  What is it exactly that I'm misunderstanding or overlooking here?
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In Munkres’ topology, there are two instances where we would have to represent the empty set as a union of sets.  First, we have a lemma that says if $X$ is a set; and $\mathcal{B}$ is a basis for a topology $\tau$ on $X$, then $\tau$ is equal to the collection of all unions of elements of $\mathcal{B}$.  Later on, we define a subbasis $S$ of a topology on a set $X$ to be a collection of subsets of $X$ whose union equals $X$, and the topology generated is the collection of all unions of finite intersections of elements of $S$.  So there are a few questions I wanted to ask about these just to make sure that I'm understanding them correctly. First in the case of the subbasis, when Munkres says finite intersections, he's allowing for a set to be intersected with itself right? There's nothing that says the sets have to be different? This is important because we will need the elements of S to be in the collection of finite intersections so that we can get the full set X.  Secondly, and more importantly, when dealing with the empty set, how do we say that this set is a union of anything? If I have a collection of sets D and I take the empty collection $\mathcal{A}$, then can I say the union of the empty collection is the empty set? I think this is ok to do, and Munkres mentions this earlier in the book, so is he allowing this to occur when he says all unions of elements from a collection of sets? I suppose its not that hard to understand but I guess I'm not used to thinking this way.  Lastly, in the proof of the lemma I mentioned first, Munkre's proves that every open set is a union of basis elements in the following way. Let $U \in \tau$, choose for each $x \in U$ an element $B_x$ of the basis such that $x \in B_x \in U$. Then $U = \bigcup_{x \in U} B_x$, so $U$ equals a union of elements of the basis.  I'm having trouble interpreting this statement if we start with $U=\emptyset \in \tau$. Then there is no $x \in U$, so how do we interpret the rest of the statement?  Thank you
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I'm working through a book on fundamental topology, and I'm stuck at a proof in set theory. The book provides a proof for:   $$\text{Let }\left\{A_\alpha\right\}_{\alpha\in I}\text{ be an indexed family of subsets of a set }S\text{; then:}$$ $$C\left(\bigcup_{\alpha\in I}A_\alpha\right)=\bigcap_{\alpha\in I}C\left(A_\alpha\right)$$  where $I$ is an indexing set for $\alpha$, and $\alpha\in I$, and $C(S)$ denotes the complement set. They do this by proving first that, if $x\in C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, then $C\left(\bigcup_{\alpha\in I}A_\alpha\right)\subset\bigcap_{\alpha\in I}C\left(A_\alpha\right)$. Then, the prove the converse; that if $x\in\bigcap_{\alpha\in I}C\left(A_\alpha\right)$, then $\bigcap_{\alpha\in I}C\left(A_\alpha\right)\subset C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, so the two sides, being subsets of each other, must be equal.  That makes perfect sense, and is very clear and straightforward. However, what I don't understand is their second proof (under the same conditions):  $$C\left(\bigcap_{\alpha\in I}A_\alpha\right)=\bigcup_{\alpha\in I}C\left(A_\alpha\right)$$  While this statement intuitively makes sense to me, I can't seem to prove it mathematically. Here's what I've done so far:  Let $x\in C\left(\bigcap_{\alpha\in I}A_\alpha\right)$. We know that, conversely, $x\notin \bigcap_{\alpha\in I}A_\alpha$. This is where the proofs diverge, though; if it were the first formula, it would be possible to say that $x\notin \bigcup_{\alpha\in I}A_\alpha$, and as a result, that $x\notin \left\{A_\beta\right\}_{\beta\in I}$. However, I can no longer do that, and approaching it from the other direction gives the same result.  Let $x\in\bigcup_{\alpha\in I}C\left(A_\alpha\right)$. We know that $x\notin\bigcup_{\alpha\in\ I}A_\alpha$, but this still doesn't get me anywhere.   How should I approach this proof? Any hints or pointers for someone new to topology?
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I'm reading Intro to Topology from Mendelson.  The entire problem statement is,  Prove that $X$ is compact if and only if for each family $\lbrace F_\alpha\rbrace_{\alpha\in I}$ of closed subsets of $X$ that has the finite intersection property, we have $\bigcap_{\alpha\in I} F_\alpha\neq\varnothing$.  My attempt at the proof is as follows:  First assume that $X$ is compact. For the sake of contradiction, suppose that there exists a family $\{ F_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ with FIP such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$. Since $X$ is compact if $\lbrace F_\alpha\rbrace_{\alpha\in I}$ any family of closed sets such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$, then there exists a finite set of indices $\lbrace \alpha_1,\dots,\alpha_n\rbrace$ such that $\bigcap\limits_{i=1}^n F_{\alpha_i}=\varnothing$. Yet, since $\lbrace F_\alpha\rbrace_{\alpha\in I}$ has the finite intersection property, for every finite index of $J\subset I$, $\bigcap\limits_{\alpha\in I} F_\alpha\neq\varnothing$.  Suppose now that $X$ is not compact, that is, there exists an open cover $\lbrace U_\alpha\rbrace_{\alpha\in I}$ of $X$ with no finite subcover. Consider the set $F_\alpha=C(U_\alpha).$ Then $\lbrace F_\alpha\rbrace_{\alpha\in I}$ is a collection of closed subsets of $X$ with the finite intersection property and were $\bigcap_{\alpha\in I} F_\alpha=\varnothing.$  Thanks for any feedback!
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Let $V$ be a finite dimensional vectorspace over a field $\mathbb{ F}$. It's easy to show that if $U$ and $V$ are subspaces of $V$ then $U \cap V$ is a subspace. But what if there are an infinite number of them? I have seen sometimes in pure maths that you take an arbitrary finite collection of sets and show the statement you are trying to prove is true for them to deduce it is true for the whole infinite collection. Why is this technique allowed? Can I do it for this case? Why?
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This is a homework question of mine that Ive answered and I'd really like some feedback on my solution.       Prove or find a counter example to the claim that for all sets $A,B,C$, $$\text{if  }A\cap B=B\cap C=A\cap C=\emptyset \text{ then } A\cap B\cap C\neq \emptyset$$.   Since I can only either prove it is true, OR find a counter example I have gone with the latter because I think this statement is not true. My counter example is as follows:  Let $A = \{1,2,3 \}$ , $B = \{4,5,6 \}$ and $C = \{7,8,9 \}$. It is then clear that $A ∩ B =  B ∩ C =  A ∩ C = \emptyset$ .  If we let $x ∈ A\cap B\cap C$ then $x$ must be in $A$ and $B$ and $C$. Since this cannot be possible because there are no common elements then this implies that $A∩B∩C$ has to be equal to a null set i.e $A∩B∩C = \emptyset$ and thereby showing that (at least for this scenario) if $A \cap  B =  B \cap  C =  A \cap  C = \emptyset$ then $A\cap B\cap C = \emptyset$.  Is this okay? Any feedback will be greatly appreciated thank you!
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Can someone please verify my proof?     Is the collection $\tau_\infty = \{U:X-U$ is infinite or empty or all of $X\}$ a topology on $X$?   No. Let $X = \mathbb{R}$.  Clearly, $\{x\} \in \tau_\infty$ for all $x \in X$.  Set $\displaystyle{B = \cup_{x \neq 0} \{x\}}$. Clearly, $X - B = \{0\}$, but $\{0\} \notin \tau_\infty$. Since $\tau_\infty$ is not closed under arbitrary unions, it does not define a topology on $X$.
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Let $(X,\tau)$ be a topological space, $U \subseteq X$ a non-empty open subset of $X$ and let $W$ be an irreducible non-empty closed subset of $X$. Assuming $W \cap U \neq \emptyset$ why is this intersection irreducible in $U$ and its closure (taken in $X$) equal $W$?  Clearly $W \cap U$ is dense in $W$ because it is an open subset of $W$ and $W$ is irreducible so its closure, taken in $W$ is $W$ itself. I don't see why its closure taken in $X$ is equal $W$. Why is this? Also, why is it irreducible in $U$?  EDIT: OK $W=cl_{X}(W)=cl_{X}(W \cap U)$ I think. Why is it irreducible in $U$ though?
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Proof:- From (c) $X\in \mathscr T$. $X\subset cl(X)[\because by (a)]$ and also we have $cl(X)\subset X$. Hence $cl(X)=X$ and $\emptyset \in \mathscr T$.   Let $\{U_i\}_{i=1}^n$ be a collection of open sets in $\mathscr T$. For each $i$ there exists $C_i$:$cl(C_i)=C_i$ and $U_i=X\setminus C_i$ then $\bigcap_{i=1}^{n}U_i=X\setminus \bigcup_{i=1}^n C_i$. We have $cl(C_i)=C_i,i=1,2,3,..,n$. By (d), we have $\bigcap_{i=1}^{n}U_i \in \mathscr T$. How do I prove arbitrary union axiom? If there is an axiom $\bigcap_{\alpha \in \Lambda}cl(A_\alpha)=cl(\bigcap_{\alpha \in \Lambda}A_\alpha)$. I could prove. Can you please help me?
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Prove the existence of uncountable family $\{X_i \subseteq \mathbb{N}\}_{i \in \mathbb{R}}$ such that intersection of any $k$ sets from this family is infinite and intersection of any $k+1$ sets is finite. $k \ge 2$, $\mathbb{N}$ means natural numbers, $\mathbb{R}$ means real numbers.
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Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.  I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.  Thank you for the help
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Theorem Any class $\mathcal{A}$ of subsets of a nonempty set $X$ is the subbase for a unique topology $\tau$ on $X$. That is to say, the finite intersections of members of $\mathcal{A}$ form a base for the topology $\tau$ on $X$. Example Consider $X=\{a,b,c,d\}$ Let $\mathcal{A}=\{\{a,b\},\{b,c\},\{d\}\}$ The finite intersection of members of $\mathcal{A}$ give us: $$\mathcal{B}=\{\{a,b\},\{b,c\},\{d\},\{b\},\varnothing,X\}$$  How is it true that $X \in \mathcal{B}$ ? By definition of subbase, the finite intersection of members of $\mathcal{A}$ are the members of some base $\mathcal{B}$. But for $a_i \in \mathcal{A}, \displaystyle\bigcap a_i \neq X$ for any finite intersection of $a_i$'s. Look: $\{a,b\}\cap\{b,c\}\cap \{d\} \neq \{a,b,c,d\}$ nor do any other intersection of elements of $\mathcal{B}$ result in $X$. Anybody know what's going on here? Why is $X \in \mathcal{B}$?
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Hi I'm new to set theory. I need to prove that if $A \subseteq B$, then $A \cap B = A$. I would like to do this the formal way, without a Venn diagram. How should I proceed?
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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.