$C$ = Cantor set. $C$ is closed, so $C^c$ is the . So let $$C^c = \bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} A_i $$ where $A_i$ are disjoint open intervals. For each $i \in \mathbb{N},\ A_i\ $ differs from a closed set by just two points, namely the endpoints of the interval $A_i.$ For each $i \in \mathbb{N},\ $ let $$ B_i = \{x_{i1}, x_{i2} \} = \{ \text{ the two endpoints of } A_i \}. $$ Then $\left|\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i\right| \leq 2 \left|\mathbb{N}\right| $ is countable. However, $\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i$ is surely the Cantor set, which is uncountable. Something is incorrect in all of this, but what?
22224
The Cantor set is closed, so its complement is open.  So the complement can be written as a countable union of disjoint open intervals.  Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?
55237
I was playing with Baire's Theorem, and seemed to deduce the following:     In a complete metric space $X$ that has no isolated points, any countable intersection of open dense sets is uncountable.   Proof: Let $S$ be the countable intersection of open dense sets $\{U_n\}$. By Baire's Theorem $S$ is dense thus $U_n$ are open and dense. This implies that $U_n^c$ are nowhere dense.  $$X = S\cup S^c = S\cup (\bigcap_n U_n)^c = S\cup \bigcup_n U_n^c.$$ If $S$ were countable, then $X = \bigcup_{s\in S}\{s\} \cup \bigcup_n U_n^c$. Since $X$ has no isolated points, every singleton set is closed and has empty interior, thus nowhere dense. We have just written $X$ has a countable union of nowhere dense sets. A contradiction to Baire's corollary.  As a corollary, any connected complete metric space, such as $\mathbb{R}$, must be uncountable (this gives a topological proof that $\mathbb{R}$ is uncountable). Also $\mathbb{Q}$ is not the intersection of open dense sets of $\mathbb{R}$.  I am a little bit suspicious about my proof, is it correct?
40867
Let $(X, d)$ be a metric space, and $(x_n)_{n \in \mathbb{N}}$ a sequence in $X$, with $x_n \rightarrow x$ w.r.t. $d$ for some $x \in X$. Show that the set $\{x_n : n \in \mathbb{N}\}\cup\{x\}$ is closed in $(X, d)$.   I figured I should use $\overline{F} = F \cup F'$ to show the above set is closed. Let $(x_n)_{n \in \mathbb{N}} = F$. Then, all elements of the sequence are in  $F$, which is equal to $\{x_n : n \in \mathbb{N}\}$. Clearly, the sequence (or $F$) has only one accumulation point, $x$. So, $F' = \{x\}$. This shows $\{x_n : n \in \mathbb{N}\}\cup\{x\} = \overline{F}$, and thus is closed.  Is this sufficient in proving the problem? Or would I have to show that $x$ is the only accumulation point as well (figured I don't have to because we know limits are unique)? Any advice would be very helpful. Thank you.
56851
Suppose that $\{A_i\}$ is a finite collection of closed sets in $X$ such that $X = \bigcup A_i$, and $f : X \to Y$ restricted to each $A_i$ is continuous. Then $f$ is continuous.   Okay. The base case ($n=2$) is just the pasting lemma. I am having a little trouble with the inductive case, specifically with what the hypothesis is and how to deal with continuity with respect to subspaces. I believe the inductive hypothesis is to assume that any function over a topological space that is the union of $n-1$ closed sets such that the function restricted to each of these $n-1$ closed set is continuous must be continuous over the whole space. Then I assume I have a topological space $X = \bigcup_{i=1}^n A_i$, where each $A_i$ is closed, and a function $f : X \to Y$ such that $f|_{A_i}$ is continuous for each $i$. Thus $X = \bigcup_{i=1}^{n-1} A_i \cup A_n$, where $C = \bigcup_{i=1}^{n-1} A_i$ is closed since it is a finite union of closed sets. If I can show $f|_C$ is continuous, then I simply apply the pasting lemma to finish the problem. Give $C$ the subspace topology, and notice that each $A_i$ is closed in $C$. Thus we have a topological space $C$ that is the union of $n-1$ closed. But do I still have continuity $f$ restricted to $A_i$ since I am no longer considering them subspaces of $X$ but of $C$? When one says $f|_{A_i}$ is continuous, is one saying that $f$ is continuous when restricted to $A_i$ and $A_i$ is given the subspace topology?  EDIT:  I think I figured out the lemma I need. Claim: Let $A$ and $C$ be subspaces of $X$ such that $A \subseteq C$. Topologize $C$ with subspace topology with respect to $X$. Then the subspace topology on $A$ as a subspace of $X$ (call it $\tau_X$) is the same as the subspace topology of $A$ as a subspace of $C$ (call it $\tau_C$).   Proof: Let $U$ be open in $X$. Then $A \cap U$ is some open set in $A$ as a subspace of $X$. Since $C$ is a subspace of $X$, $C \cap U$ is open in $C$ as a subspace of $X$. Then $A \cap (C \cap U) = (A \cap C) \cap U = A \cap U$ is open in $A$ as a subspace of $C$, thereby proving $\tau_X \subseteq \tau_C$. Now let $O$ be some open set in $C$ as a subspace of $X$. Then $A \cap O$ is some open subset of $A$ as a subspace of $C$. But $O = C \cap U$ for some open set in $X$. Therefore $A \cap O = A \cap (C \cap U)= A \cap U$, which is also open in $A$ as a subspace of $X$. Hence $\tau_C \subseteq \tau_X$  Kind of obvious. But it was hard to spot the need for this lemma when I was considering subspace of subspaces and continuous functions, etc. Am I right in thinking this is the lemma I need?
15311
Claim: If $(X,\tau)$ is not $T_1$ then $A'$ is not necessarily closed for any $A$.  Proof: Here is an example. Take $X=\mathbb{N}$ and let $\tau=\lbrace u\subset\mathbb{N}\mid 1\in u\iff 2\in u\rbrace$. Let $A=\lbrace 2,3 \rbrace$. Since any open set containing $1$ also contains $2$ it follows that $1\in A'$. Hence $\overline{A}=\lbrace1,2,3\rbrace$, as it is the smallest closed set containing $A$. But  $X\setminus A'=\mathbb{N}\setminus\lbrace 1\rbrace$ is not open because it contains $2$ and not $1$. The set of accumulation points $A'$ is not closed.  So, in general, trying to prove that $A'$ is closed in a topological space is an effort in futility?
36217
I thought I had the proof of this in my head, but it doesn't sound right on paper. Can someone see if my argument could be improved.  Let $(X,\tau)$ be a topological space that is separable, then it contains a countable dense subset $\{D_n\}_{n \in \mathbb{N}}$. Then $X$ is CCC if it does not contain any uncountable collection of disjoint open sets.  My argument:      By contradiction, suppose $X$ is separable but is not CCC.       Since a countable dense subset could only intersection countably many   open sets, there exists some open sets $U$ that $D_n$ fails to   intersection.      Therefore, $X$ is not separable as claimed.    My problem with this statement is that: Since a countable dense subset could only intersection countably many open sets...  Is there to make the middle paragraph rigorous? The whole proof feels wrong, please assist!
108507
I don't understand a proof right at the beginning of this document found .   This proof is on why any base for the open sets in a second countable space has a countable subfamily that is a base.  My trouble comes in the second paragraph of the first proof. It says $\{B_{i_x}\mid x\in C_k\}$ is a countable set whose union is $C_i$. First, why is the set countable? If you take a set $B_{i_x}$ for any $x\in C_k$, what if $C_k$ is not countable itself? And why is the union $C_i$? Should it instead be $C_k$? I don't see $C_i$ mentioned anywhere in the second paragraph.
20353
There was an exercise I could not do. So the property is $T_1$ if for every pair of distinct points, $P, Q \in X$, there is an open subset $U$ containing $P$ but not $Q$ and another open subset $V$ containing $Q$ but not $P$.  So I was asked to show if the base field is infinite, then $\mathbb{A}^n$ is $T_1$ but not Hausdorff.  I can show that it's not Hausdorff (I believe this is right). So let $k$ be an infinite field, then we have $U$ and $V$ which are two nonempty subsets which are open, thus $U^c$ and $V^c$ finite so by Demorgan's law we have $(U \cap V)^c =U^c \cup V^c$ which is also finite, so this implies that $U\cap V$ is non-empty so it can't be Hausdorff.  But I can't seem to really show that it's $T_1$  Any ideas how to approach this?  Thank you.
87046
Let X be an uncountable set. And let $$\mathcal{A} =\{A\subset X: A \text{ is countable or } X-A \text{ is countable}\}.$$ Then it can be shown that $\mathcal{A}$ is a $\sigma$-algebra. Define $\mu:\mathcal{A}\rightarrow [0,\infty]$ by $\mu(A)=0$ if $A$ is countable or 1 otherwise. The task is to show that $\mu$ so defined is a measure on $\mathcal{A}$.  It is clear that $\mu(\emptyset)=0$. Now we want to show that $\mu$ is countably additive on disjoint sets from $\mathcal{A}$. I'm first trying to show this for two disjoint sets $A$ and $B$ in $\mathcal{A}$. I am running into a problem if I assume that both $X-A$ and $X-B$ is countable.  For if this happens then, $B\subset X-A$ implies $B$ is countable. But both $B$ and $X-B$ cannot be countable, since then $X=B\cup (X-B)$ will be countable, running contrary to the assumption that $X$ is uncountable. How should I address this case?
91712
Cantor defined an infinite set as a set whose subset can be placed in a one-to-one correspondence with its subset. That is, take the set of all natural numbers: {0, 1, 2, 3, 4,...}. From that set, you can form a subset of all even numbers: {0, 2, 4, 6,...}. There is a one-to-one correspondence between both these sets. After all, you can multiply each member in the set of all natural numbers by 2, and you’ll get the set of all even numbers. Because of this one-to-one correspondence, the two sets have the same cardinality, despite the set of even numbers being a subset of the set of natural numbers. This is counter-intuitive to me. How can a subset of a set be as big as the set itself? To make the intuition more concrete, consider the set of multiples of 1,000,000: {0, 1,000,000, 2,000,000,...}. There is a one-to-one correspondence between this set and the set of all natural numbers. So they have the same cardinality. This is really counter-intuitive.  From as far as I can tell, there are basically two axioms at play here:  A1: If all members of set A can be placed in a one-to-one correspondence with all the members of set B, then both sets have the same cardinality.  A2: If set A is a proper subset of set B, then A has a smaller cardinality than set B.  How Cantor defined infinite sets is that he accepted A1 and rejected A2. That is counter-intuitive to me. But mathematicians might say that my intuition is just wrong. So my question is whether we can define infinite sets in another way, i.e. accept A2 but reject A1. That is, the set of even numbers is smaller than the set of all natural numbers, even though both sets have a one-to-one correspondence. If you think this is counter-intuitive because sets with one-to-one correspondence with each other should have the same cardinality, and thus think this is wrong, why can’t I say the same when Cantor accepted A1 and rejected A2? Either way, there is something counter-intuitive and paradoxical.   So, my questions are:   Can we accept A2 and reject A1 for the definition of an infinite set? What arguments, if any, are given for accepting A1 and rejecting A2?
71765
I would be very glad if you helped me solving this problem. It's about topology. Okay let's begin :    Let $A$ be a non-empty open subset of $\mathbb R$. Let $x_0 \in A$ and $B = [x_0,\infty) ∩ A^c$ (complement of $A$)  First I should prove that $B$ is a non-empty, closed set. Then I should prove that :     $B$ is bounded below.   $\inf B &gt; x_0$   $\inf B$ doesn't belong to $A$.   $[x_0, \inf B) \subseteq A$.    Then I should explain how to deﬁne a number $\beta$ so that $(\beta,\inf B) \subseteq  A$, where $\beta$ doesn't belong to $A$.    I really can't get up with proofs for these question. Can you please help me solving this or at least give me some hint or sketch? I would be very grateful. Thank you :)
33208
Let $C$ be a chain of countable sets, i.e. $\forall S, T \in C: S \subseteq T \lor T \subseteq S$, and that every $S \in C$ is countable. Then, is $\bigcup C := \{ t \mid \exists S \in C: t \in S\}$ countable?  The answer is no, and a counter-example is $C = \omega_1$, the first uncountable cardinal/ordinal. Is there a more elementary example that doesn't involve, say, cardinals and ordinals?
53345
Proposition : The only sets in $R^n$ which are both open and closed are the empty set and $R^n$ itself.  I came up with the proof of this claim and I'd like to know if my proof is correct.  Proof:  Let $S$ be the nonempty proper subset of $R^n$ which is both open and closed. Then $T = R^n-S$ is also a nonempty proper subset of $R^n$ which is both open and closed. Now take a point $s_0$ in $S$ and $t_0$ in $T$. Then we can consider a closed box $J_1$ which contains both $s_0$ and $t_0$.  The n-dimensional interval $I_1^k$ is defined by the inequalities   $s_k\le x_k \le t_k$ (k=1, 2, $...$, n ), where each $s_k$ and $t_k$ are the k-th coordinates of $s_0$ and $t_0$. Without loss of generality, assume that for all k, $s_k \le t_k$.  Here $J_1$ denotes the catesian product   $J_1$ = $I_1^1$ x  $\cdots$ x $I_n^1$   Now, bisect all of the n 1-dimensional intervals and the point $\frac {s_0+t_0}{2}$ will lie in the center of the bisected box. This point belongs to either $S$ or $T$. Say it belongs to $S$ then, rename it $s_1$ and the original $t_0$ to be $t_1$. Now again consider a box that has both points as vertices like the previous case and repeat the process.   Writing $I_k^m$ = [$s_k^m$,$t_k^m$],   we have  $t_k^m$ - $s_k^m$ = $\frac {t_k-s_k}{2^{m-2}}$  Then for each fixed k, the sup of all left endpoints must therefore be equal to the inf of all right endpoints, and their common value we donote by $b_k$ (actually we know this by Bolzano-Weierstrass).   Now we assert that b=($b_1$,$...$,$b_n$) is a limit point of $S$ and $T$. To see this, take any n-ball B(b; $r$). The point b of course belongs to each of the boxes $J_1$, $J_2$, $...$ constructed above, and when $m$ is such that $max_k\frac {t_k-s_k}{2^{m-2}}$ &lt; r/2, this neighborhood will include $J_m$ and since $J_m$ contains both points of $S$ and $T$, b is a limit point of both sets. But since both sets are closed, they both contain b, which is a contradiction. Hence, the proposition is true.
44326
I am trying to prove that $\mathbb{R}$ with the lower limit topology is not second-countable.  To do this, I'm trying to form an uncountable union $A$ of disjoint, half-open intervals of the form $[a, b)$, $a &lt; b$. Is this possible? I think this would imply the $A$ is open but no countable union of basis elements could coincide with $A$ therefore making the real numbers with the lower limit topology not second-countable.  I think there must exist something like $A$ described above but I am having trouble visualizing it and coming up with a formula to represent it.  Maybe there is some other way to show it is not second-countable.
40613
I recently started studying set theory and am having quite a bit of difficulty accepting Cantor's diagonal proof for the uncountability of the reals. I also saw a topological proof via nested sets for uncountability which still does not satisfy me completely, given that just like the diagonal it relies on a never ending process. In fact, the nested sets proof sounds very much like the diagonalization proof to me.  Do all proofs of the uncountability of the reals involve diagonalization? Are there any other proofs I can look at to understand? I couldn't find any on searching stack exchange. Thanks.
106871
Following a reference from "General Topology" by Stephen Willard   Well since the difference between an open set and a closed set is an open set and since the union of open sets is an open set, it is clearly that $S$ and $T$ are open; but unfortunately I can't prove that $S$ and $T$ are disjoint open set containing $A$ and $B$. Anyway it seems to me that $U=\bigcup_{i=1}^nU_i$ and $V=\bigcup_{i=1}^nV_i$ are two disjoint open sets that containing $A$ and $B$ and so I ask: why define the set $S$ and $T$? Could someone help me?
49228
If we say that $\mathcal{S}$ is a family of disjoint triangles on a plane, is $\mathcal{S}$ a countable or an uncountable set?  I believe that it is an uncountable set since $\mathbb{R}$ is uncountable and our triangles' corner coordinates are $\in \mathbb{R}$, but I cannot prove this formally, any tip would help.