Let $\{T_\alpha\}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $T_\alpha$, and a unique largest topology contained in all $T_\alpha$.   We can check that $\bigcap T_\alpha$ is a topology on $X$, so it is the unique largest topology contained in all $T_\alpha$.   Now, a topology containing all $T_\alpha$ must contain $\bigcup T_\alpha$. It must also contain arbitrary unions and finite intersections of sets in $\bigcup T_\alpha$. Since the union of the sets in $T_\alpha$ is $X$, this is the topology generated by the subbasis $\bigcup T_\alpha$. How can we prove that it is the unique smallest one containing all the collections $T_\alpha$?
63151
I'm doing this exercise:     Let $\{T_\alpha\}$ be a family of topologies on $X$. Show that there   is a unique smallest topology on $X$ containing all the collections   $T_\alpha$, and a unique largest topology contained in all $T_\alpha$.   I have proved everything except the unique part.. I just can't get my head around what is meant with unique here. Which may sounds silly.   I have proved that the intersection is a topology. And if you are a topology that is also contained in every $T_\alpha$, than you surely are contained in the intersection, so you are not larger.  But I don't see from what it follows that this intersection is the unique largest topology contained in all $T_\alpha$. One part of my head say it is trivial, the other part gets confused. Like it is redundant to talk about unique in this context.  The same for proving the uniqueness of the smallest topology.   Edit Should I read topology $A$ larger than topology $B$ as, $A$ has more elements than $B$ ? I thought that, because the author uses the word finer for $A \supset B$.
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Show that the dictionary order topology on $\mathbb R\times \mathbb R$ is the same as the product topology $\mathbb R_d\times \mathbb R$ where $\mathbb R_d$ is $\mathbb R$ with the discrete topology.  My thoughts:  First, the space $\mathbb R_d\times \mathbb R$ has $\{x\}\times (a,b)$ as its basis. In particular, any set of the form $\{x\}\times (-\infty,+\infty)$ is open there, since $$\{x\}\times (-\infty,+\infty)=\bigcup_{r\in \mathbb R_{&lt;a}} \{x\}\times (r,b)\bigcup_{s\in \mathbb R_{&gt; b}} \{x\}\times (a,s)$$ Second, the open sets in $\mathbb R\times\mathbb R$ in the dictionary order topology is one of the two forms:   Consider an open set in the dictionary order topology. If it has the first form, then it is open in $\mathbb R_d\times \mathbb R$ because it is of the form $$\bigcup_{r\in \mathbb R_{&gt; a}}\{a\}\times (a,r)\bigcup_{t\in (a,b)_{\mathbb R}}\{t\}\times (-\infty,+\infty)\bigcup_{s\in\mathbb R_{&lt; d}} \{b\}\times (s,d).$$ A set of the second form is clearly open in the product topology. Thus the dictionary order topology is contained in the product topology.  Is the above correct?  The converse is somewhat unclear. If we have a set open in the product topology, it does not necessarily have one of the two forms above. It may have the form $\{x\}\times (-\infty,y_0)$ or $\bigcup_{x\in [r,s]_{\mathbb R}}\{x\}\times (-\infty,y_0)=[r,s]\times (-\infty,y_0)$. What to do in such cases?    I've just realized that the sets in the picture form a basis, they are not all open sets. Then I it suffices to show that any basis element of the product topology, namely $\{x\}\times (a,b)$, is a subset of a basis element of the dictionary order topology, and conversely, right? The former is obvious. The latter is obvious for basis elements of the second type. But it's still unclear why a basis element of the first type is contained in a set of the form $\{x\}\times (a,b)$.
54095
Question: Let $(X, d)$ be a metric space. Let $A$ and $B$ be disjoint subsets of $X$ that are closed in the topology induced by $d$. Prove that there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B\subset V$.   Is this proof correct? : By assumption, $A\cap B=\emptyset$, so there is no pair like $(x_A,x_B)$ (i.e., $x_A\in A$ and $x_B\in B$), such that $x_A=x_B$. Because every metric space is Hausdorff, so for every pair $(x_A,x_B)$ there are two disjoint open sets $U_A\ni x_A$ and $V_B\ni x_B$; and thus union of all $U_A$s (which is an open set containing A) are disjoint from union of all $V_B$s (which is an open set containing A).   Thanks a lot for any help.    PS In special case of standard metric topology on $\mathbb{R}$, if $[a,b]$ and $[c,d]$ are disjoint (say $c&gt;b$), intuitively it is apparent that there is a 'space' between $c$ and $b$ so we can have two disjoint open sets, each of them containing $[a,b]$ or $[c,d]$.
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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'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.  "Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."  Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.  Thanks.
45203
We have the classic result that an open subset of $\mathbb{R}$ is an at-most-countable union of disjoint open intervals.  In trying to prove this, I found the following line of reasoning, and I would like to know if its correct, because it's quite different from the other proofs I have read.    Let S be an open subset of R.  Thus for every $a$ in $S$, $a$ is contained in some open ball $B$ which is contained in $S$.  In particular $a$ is contained in an open interval which is contained in $S$.  For each $a$, let's associate the maximal such interval $I_a=(\alpha, \beta)$ where $\alpha$ and $\beta$ live in $\mathbb{R} \bigcup \{\pm\infty\}$  $\alpha=\inf\{y: (y,a] \subset S\}$  $\beta=\sup\{y: [a,y) \subset S\}$  Now we have that $S$ is the union of $I_a$ over all $a$ in S. Furthermore if $I_{a_1}$ and $I_{a_2}$ are distinct, then they must be disjoint (otherwise we would contradict the maximality of at least one of them).   Hence $S$ is a disjoint union of open intervals.   Finally, each such interval contains a rational number which is not contained in any other such interval; if there were uncountably many intervals in the union, there would be uncountably many rationals, a contradiction.  Thus $S$ is an at-most-countable union of disjoint open intervals.
20416
Let $X$ be a topological space containing two closed points $x,y$ and let $i : \{x,y\} \to X$ denote the inclusion map. Notice that $\{x,y\}$ carries the discrete topology. Let $F$ be a sheaf on $X$. Then $i^{-1} F$ is a presheaf on $\{x,y\}$ which is given by $(i^{-1} F)(\emptyset)=1$ (the terminal set), $(i^{-1} F)(\{x\}) = F_x$ (the stalk at $x$), $(i^{-1} F)(\{y\})=F_y$ (the stalk at $y$) and $$(i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).$$ Why is, $i^{-1} F$ a sheaf if and only if the canonical map $$\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.$$ is an isomorphism ?.   Thanks a lot to all of you.
56899
To say that $p$ is a condensation point of a subset $S$ of a topological space $X$ is to say that any open neighbourhood of $p$ contains uncountably many points of $S$.  Let us suppose we have written $\mathbb{R} = \bigcup_{i=1}^\infty A_i$. I would like to know whether the condensation points of some $A_i$ must form a set with nonempty interior.  My thoughts: First of all, some of the $A_i$ may be countable. But, the union of all such $A_i$ is 1st category in the Baire space $\mathbb{R}$, so the union of all the uncountable $A_i$ is 2nd category in $\mathbb{R}$. Hence there is some uncountable $A_n$ which is not nowhere dense. The difficulty is that we could have $A_n$ equal to, say, the union of the rationals with a Cantor set, in which case the condensation points of $A_n$ are exactly the Cantor set (which has empty interior).   It is my feeling, although I can't seem to prove it, is that there must be at least one $A_i$ out there which makes use of its uncountability to fail to be nowhere dense.    I think the answer to this question is "yes". The key thing I hadn't noticed which makes this easy, pointed by Bryan Scott below, is that if $A \subset \mathbb{R}$ and $C$ is the set of condensation points of $A$, then $A - C$ is countable. As I see it, this is because of the following  Fact: If $S$ is an uncountable subset of a second-countable topological space $X$, then $S$ has a condensation point.  Proof. Otherwise, there is a cover $(U_s)_{s \in S}$ of $S$ such that each $U_s$ contains only countably many points of $S$. Since $X$ is second-countable, $S$ is second-countable, hence Lindelof in the subspace topology. Thus we can extract a countable subcover from $(U_s)$ and this shows, contrary to assumption, that $S$ is countable. QED.  Continuing, if $A-C$ were uncountable, it would have a condensation point which would also be a condensation point of $A$ so $A-C$ must be countable. Since countable implies 1st category, it follows from $\mathbb{R} = \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n (A_i-C_i) \cup C_i$ that for some $i$ the condensation points $C_i$ of $A_i$ do not form a nowhere dense set. Since the condensation points of a subset of a topological space always form a closed set (clearly their complement is open), this $C_i$ is closed and not nowhere dense, or equivalently, is closed with nonempty interior.
95619
Consider Theorem 19.1 from Munkres' topology:     The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.   I am having a hard time seeing how the last sentence follows from the subbasis-definition of the product topology. I don't find Munkres' preceding explanation to be helpful.
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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.
118477
Most topology texts go on directly to give definition of topology, then they give some examples and that's it, like they directly tell you right     Let $X$ be a set and let $τ$ be a family of subsets of $X$. Then $τ$ is called a topology on $X$ if:         Both the empty set and $X$ are elements of $τ$   Any union of elements of $τ$ is an element of $τ$   Any intersection of finitely many elements of $τ$ is an element of $τ$      But why did we define it this way? What's the intuition?   Does anyone know a good text on topology which gives the intution behind the concepts, ...etc ?
22194
I have almost solved the following problem but am stuck at the very end, can you help me finish it? Thank you for your help.       Let $n&lt;\omega$ and $t\in {}^n\omega$. We define $U_t=\{s\in {}^\omega\omega : t\subseteq s\}$. The family   $\mathcal B=\{U_t : t\in\bigcup {}^n\omega\}$ is a basis for a topology $\tau$ on ${}^\omega\omega$. This means that a set $X$   is in $\tau$ if and only if $X$ is a union of elements of $\mathcal B$.      EXERCISE 50 (PG): Show that $\langle {}^\omega\omega,\tau \rangle$ is a second countable completely metrizable space. Hint:    For $r,s\in {}^\omega\omega$ such that $r\ne s$, the $\varrho(r,s)=1/\min\{n : r(n)\ne s(n)\}$. Moreover, let $\varrho(r,r)=0$.    Show that $\varrho$ is a complete metric on ${}^\omega\omega$ that induces the topology $\tau$.     (i) Second countable: the set of $n$-tuples is countable since it is a finite union of countable sets.  (ii) complete with respect to $$ d(f,g) = \frac{1}{\min \{n \mid f(n) \neq g(n) \}}$$ Let $f_k$ be a Cauchy sequence. This means that the index at which $f_k, f_i$ differ gets larger as $i,k$ get larger. Let $f(n)$ denote the pointwise limit, that is, $\lim_{k \to \infty} f_k(n)$. Then $f(n) \in \omega$ for every $n$ since $f_k(n)$ is eventually constant. Hence $f \in \omega^\omega$.  (iii) $d$ induces the same topolgy as $U_t$:  $\tau_t \subset \tau_d$: Let $s \in U_t$. Then $B(s, \frac{1}{n+1})$ is an open ball contained in $U_t$ hence $U_t$ is open in $\tau_d$.  $\tau_t \supset \tau_d$: Let $B(s, \frac{1}{k})$ be an open balls. If $k &lt; n$, $s \in U_s \subset B(s, \frac{1}{k})$. But if $k \ge n$ then I'm stuck. The ball $B(s, \frac{1}{k})$ consists of all sequences that agree with $s$ on the first $k$ coordinates. So it is quite small. How do I make the $U_t$ small enough to fit in this ball? I have thought of intersection but since I'm only allowed to do finite intersections I don't see what to do.
54218
This problem is from John Lee's "Introduction to Smooth Manifolds"  1-4. Let M be a topological manifold, and let U be an open cover of M .  (a) Assuming that each set in U intersects only finitely many others, show that U is locally finite.  (b) Give an example to show that the converse to (a) may be false.  (c) Now assume that the sets in U are pre compact in M; and prove the con- verse: if U is locally finite, then each set in U intersects only finitely many others.  I think I have been able to successfully proof part a). I am stuck on part b) and c).  Any help is appreciated!
101302
Exercise: What is the closure in $\mathbb{R}$of each of the following set: i) the set  $\mathbb{P}$ of all irrational numbers.  I have no idea on how to solve this question.  Definition: Let $\tau$ be a topology on a set $X$ and consider $A\subseteq X$. If $x\in X$ is an accumulation point of $A$ if: $\forall \mathscr{U}\in\tau,x\in \mathscr{U}\implies A\cap\mathscr{U}$ contains another point other than $x$.  However I don not know how to proceed using the  definition. Question: Can someone show me how to solve this question? Thanks in advance!
85634
I dont have a clue about the proof in the reverse implication on the iff statement.  I will show what I have by now:  1) A cofinite topology implies that any infinite subset in $X$ is dense. Let $U\in\tau$ and is cofinite topology so $U=X-\bigcup_{k=1}^{n}\{x_k\}$.  Let $V\in X$ an infinite subset then $$V\cap U=V\cap(X-\bigcup_{k=1}^{n}\{x_k\})=(V\cap X)-(V\cap \bigcup_{k=1}^{n}\{x_k\})=V-(V\cap \bigcup_{k=1}^{n}\{x_k\})$$  and cause $|V|\ge\aleph_0$ and $|\bigcup_{k=1}^{n}\{x_k\}|=n\in\Bbb N\,$ we have that $|V\cap \bigcup_{k=1}^{n}\{x_k\}|\le n$. And finally is obvious that $V-(V\cap \bigcup_{k=1}^{n}\{x_k\})\ne\emptyset$ so any infinite subset is dense on $X$.  2) A cofinite topology implies a $T_1$-space. Let $x, y\in X$ and $U\in\tau: U=X-\{y\}$ so $x\in U\land y\notin U$. We can define too $V\in\tau:V=X-\{x\}$ so $x\notin V\land y\in V$ then a cofinite topology is a $T_1$-space.  But my problem is that I dont have a clue to prove the reverse implication i.e. if we have a $T_1$-space where every infinite subset is dense then it is a cofinite topology. Some hint? Some advice or correction in the proof of above? Thank you in advance.    I will complete the proof with the advice of @Cameron Buie.  1) A $T_1$-space implies that any finite subset is closed. Let the singleton $\{x\}\subset X$. Then $\overline{\{x\}}=\{x\}\cup\{x\}'$ where $\{x\}'=\{y:(\forall U\in\tau : y\in U)\land x\in U\}$, but because we are in a $T_1$-space exist some $U$ for any points $x,y\in X: x\in U\land y\notin U$ so $\{x\}'=\emptyset$ then $\overline{\{x\}}=\{x\}$ and is closed.  This means that for every $T_1$-space it cofinite sets are open.  2) If any infinite subset is dense on $X$ it means that the closure of any infinite subset is $X$ so the only closed sets are finite (and $X$ itself), so the topology is cofinite.
62351
I am solving question $8$ of exercises in section 1.1. of chapter 1 in "Topology without tears". The question reads as follows.     Let $X$ be an infinite set and $\tau$ be a topology on $X$. If every infinite subset of $X$ is in $\tau$, prove that $\tau$ is the discrete topology   My idea was to show that every singleton set is in the topology and thereby the topology is discrete.  Choose an infinite set $A$ from $X$ such that $A^c$ is also infinite. $A$ and $A^c$ belong to $\tau$ since they are infinite sets. Now consider any $x \in X$. Note that $A \cup \{x\}$ and $A^c \cup \{x\}$ belong to $\tau$ since they are infinite sets. Hence, their intersection which is nothing but $\{x\} \in \tau$ for every $x \in X$.  The problem I have is I don't know how to prove the first sentence in the previous paragraph. These are my line of thoughts to prove them.   If $X$ is a countably infinite set, then I can list the elements are let the odd numbered elements fall into $A$. This guarantees $A$ and $A^c$ are both infinite. If $X$ is uncountable, then I can choose a countably infinite subset and call it $A$.   Are the above arguments rigorous? I am not especially happy with my second argument of choosing a countably infinite subset from an uncountable set since I do not give an explicit procedure of constructing the set $A$.  Also, is there any other simpler way of answering the original question?
42247
Let, $\tau_1, \tau_2,\tau_3$ be three topologies on a set $X$ such that $\tau_1 \subset\tau_2\subset\tau_3$ and $(X,\tau_2)$ be compact $T_2$ space. Then,   (A) $\tau_1=\tau_2$ , if $(X,\tau_1)$ is $T_2$.  (B) $\tau_1=\tau_2$ , if $(X,\tau_1)$ is compact.  (C) $\tau_1=\tau_3$ , if $(X,\tau_3)$ is $T_2$.  (D) $\tau_2=\tau_3$ , if $(X,\tau_3)$ is compact.     I only know that closed subspace of a compact space is compact and $T_2$-ness is a heriditary property. But I have no idea how it depends on topologies , whether the topologies are finer or coarser.   Please help...........