Def: The product topology is generated by sets of the form $\prod\limits_{n\in\mathbb{N}} U_n$ where each $U_n$ is open in $X_n$ and, for all but finitely many $n$, we have $U_n = X_n$.  I am looking for an answer that explains how the open sets in the final topology is affected by the definition especially for all but finitely many $n$, we have $U_n = X_n$.
121618
Why are box topology and product topology different on infinite products of topological spaces ?  I'm reading Munkres's  topology. He mentioned that fact but I can't see why it's true that they are different on infinite products.   So , Can any one please tell me why aren't they the same on infinite products of topological spaces ?
101275
Question: Why is the topological space $\mathbb{R}^\infty$ defined to be the subset of $\prod_{i=1}^\infty \mathbb{R}_i$ consisting of sequences $(a_i)_{i=1} ^{\infty}$ such at most finitely many $a_i\neq 0$? Why does one insist on the condition that $a_i\neq0$ for at most finitely many $i$?
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?
90805
The axioms of a topological space are usually stated in the "open set" form: A topological space $X$ has a set of subsets $\tau$ whose members satisfy:   $\emptyset$ and $X$ are in $\tau$. $\tau$ is closed under arbitrary unions. $\tau$ is closed under finite intersections.   We will call this the usual open set axiomatization of a topological space. The open set axiomatization is probably the most economical, but of course there are other ways as well, such as the  which axiomatizes the topology using the closure operator.   In his note , Pete L. Clark gives several alternative characterizations of a topology, most of which have details given. It was mentioned, I think originally by Willard, that it is "possible, but unrewarding to characterize a topology completely by its [boundary]". I could find no further references on this, and I am wondering how it might be done (unrewarding as it may be).  So my question is this: How can a topological space by axiomatized using the boundary operation as the primitive notion? A complete answer should give a list of axioms required for the formulation, as well as a sketch of how they are equivalent to the usual open set formulation.
114856
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 $\{x\}\times (a,b)$.
16420
My attempt to prove the following is below:     Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$   My book calls this set the ""  I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly)  Using the definition of topology:     $\mathscr{T}$ is a topology on X if and only if the following are true:   (i) X and Ø are elements of $\mathscr{T}$.   (ii) $\mathscr{T}$ is closed under finite intersections.   (iii) $\mathscr{T}$ is closed under arbitrary unions.   Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2...$ are all finite sets)  So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.   Now to verify the definition.     (i) X and Ø are elements of $\mathscr{T}$.   Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.      (ii) $\mathscr{T}$ is closed under finite intersections.   $U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| &lt; |\mathbb{N}|  $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.       (iii) $\mathscr{T}$ is closed under arbitrary unions.   $U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| &lt; |\mathbb{N}|  $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.  Now let's consider the infinite case: $V_k \cap ... \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap ... $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.
46334
Let $\lbrace \tau_n:n\in \mathbb{N}\rbrace$ be an ascending chain of topologies on a nonempty set $X$. Then is $\bigcup\limits^{\infty}_{n=0}\tau_n$ a topology on $X$?   I have a strong notion that it may not be a topology. But I am not getting the proper counter example. Please help. I know this must be a topology if the collection is finite.
86706
I have a question about the following counterexample, why  $U_n = \{n\} \times Y$ has not finite collection covering $X$?   Is it because if you take a finite set $S\subset U_n$ then it's impossible to have $X\subset \bigcup S$ since X is infinite?  Counterexample:  Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$
113085
Let $\mathcal{U}$ be an open cover of $\mathbb{R}$ (Standard Topology) such that $\mathbb{R} \not \in \mathcal{U}$ and for any finite set $A$ there is a $U \in \mathcal{U}$ such that $A \subseteq U$. We call such an open cover an $\omega$-cover. Can we show that for any finite set $B \subset \mathcal{U}$, there is a $V \in \mathcal{U}$ such that $\cup B \subseteq V$?  Ultimately I'm working on showing the following. Let $\langle \mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of $\omega$-covers. Can we find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each $F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an open cover of $\mathbb{R}$?  My approach here was to use each $\mathcal{U}_n$ to cover $[-n,n]$, thus eventually covering all of $\mathbb{R}$. Since $[-n,n]$ is compact and $\mathcal{U}_n$ is a cover, $\mathcal{U}_n$ has a finite subcover. But that's as far as I can get unless what I conjectured above is true.
72699
I see this everywhere, but cannot find a proof of it. Is it just this easy; it feels wrong.  Suppose $X$ is noet and $Y\subset X$ with subspace topology. If $Y$ has a descending chain of closed sets $Y_1 \supset Y_2 \supset Y_3 \supset \ldots$, where each $Y_i$ is formed by intersection some closed set in $X$ with $Y$. That is $(X_1\cap Y)\supset (X_2\cap Y) \supset \ldots$ but I can't seem to finish this because it is not as if I can just remove the intersections with $Y$ right?
112574
Let $\{X_\alpha\}_{\alpha\in J}$ be a family of topological spaces and let $X=\prod_{\alpha\in J}X_\alpha$ be endowed with the product topology. Show that $X$ is path connected if and only if each $X_\alpha$ is path connected.  I know this is true if path coennected is changed by connection but I am very confused with this problem, I have tried to do the following:  $\Rightarrow $ let $x,y \in X_\alpha$ be then $(x_\alpha), (y_\alpha)\in \prod X_\alpha$ where $x_\alpha=x, y_\alpha=y$ and $x_\beta=0, y_\beta=0$ for any $\beta\neq\alpha$, and since P is path connected then there is continuous $f:[0,1]\rightarrow \prod X_\alpha$ such that $f(0)=(x_\alpha)$ and $f(1)=(y_\alpha)$, then if $\pi_\alpha:\prod X_\alpha \rightarrow X_\alpha$ is the projection, which is continuous, we have that $\pi_\alpha f:[0,1]\rightarrow X_\alpha$ is continuous and $\pi_\alpha (f(0))=x$ and $\pi_\alpha (f(1))=y$.  $\Leftarrow$ I do not know how to do this address, could someone help me please? Thank you very much.
98567
I know the Cantor set probably comes up in homework questions all the time but I didn't find many clues - that I understood at least. I am, for a homework problem, supposed to show that the Cantor set is homeomorphic to the infinite product (I am assuming countably infinite?) of $\{0,1\}$ with itself. So members of this two-point space(?) are things like $(0,0,0,1)$ and $(0,1,1,1,1,1,1)$, etc. Firstly, I think that a homeomorphism (the 'topological isomorhism') is a mapping between two topologies (for the Cantor sets which topology is this? discrete?) that have continuous, bijective functions. So I am pretty lost and don't even know what more to say! :( I have seen something like this in reading some texts, something about $$f: \sum_{i=1}^{+\infty}\,\frac{a_i}{3^i} \mapsto \sum_{i=1}^{+\infty}\,\frac{a_i}{2^{i+1}} ,$$ for $a_i = 0,2$. But in some ways this seems to be a 'complement' of what I need.... Apparently I am to use ternary numbers represented using only $0$'s and $1$'s in; for example, $0.a_1\,a_2\,\ldots = 0.01011101$? Thanks much for any help starting out!  Here is the verbatim homework question:  The standard measure on the Cantor set is given by the Cantor $\phi$ function which is constant on missing thirds and dyadic on ternary rationals. Show the Cantor set is homeomorphic to the infinite product of $\{0,1\}$ with itself. How should we topologize this product? (Hint: this product is the same as the set of all infinite binary sequences) Fix a binary $n$-tuple $(a_1,\ldots, a_n)$ (for e.g., $(0,1,1,0,0,0)$ if $n = 6$). Show that the Cantor measure of points ($b_k$) with $b_k=a_k$ for $k \leq n$ and $b_k \in \{0,1\}$ arbitrary for $k&gt;n$, is exactly $1/2^n$. These are called cylinders. (They are the open sets, but also closed!)
37976
(1) If $X$ and $Y$ are two sets, we define the Cartesian product $X \times Y$ as the set of ordered pairs $(x,y)$, such that $x \in X$ and $y \in Y$.  (2) On the other hand [Folland, Real Analysis, page 4], if $\{X_\alpha\}_{\alpha \in A}$ is infinite indexed family of sets, their Cartesian product $$ \prod_{\alpha \in A}X_\alpha $$ is defined as the set of maps $f: A \to \bigcup\limits_{\alpha \in A} X_\alpha$ such that $f(\alpha) \in X_\alpha$ for every $\alpha \in A$.  After saying this, Folland remarks:     it should be noted, and promptly forgotten, that when $A = \{1,2\}$, the previous definition  of $X_1 \times X_2$ [that's (1) above] is set-theoretically different from the present definition of $\prod_1^2 X_j$ [that's (2) above]. Indeed, the latter concept depends on the mappings, which are defined in terms of the former one.   I am not grasping this remark. Specifically, here are my questions.  Question 1: How is (2) set-theoretically different from (1)? A simple illustrative example?  Question 2: If (1) is extended to infinite families, which definition would be stronger? A simple illustrative example?  Question 3: Why should this be "promptly forgotten"?  I'll probably have more questions depending on the type of answers I'll get to these.  Thanks!
81111
I would like to know if there is like a magical formula to know how many topologies exist on a finite set  For example for $X = \{ a, b, c \}$ I found $29$, but I dont know if there are more or how to know this exact number without writing all topologies first.
81578
One can see on every text (book, lesson, comments) that a direct sum/coproduct of abelian groups is the same as a finite product but in the infinite case, the direct sum/coproduct is only a subgroup of the product.  Why does an element in an infinite direct sum only have finitely many components different from the neutral elements?  I don't see where the "finite" comes from when I look at the universal property.  Also, in another situation, a condition of "finiteness" appears and I think that is because of the same structure: a freely generated group, vectors space, algebra etc... has elements that are finite product/sum of generating elements. Because, if one really wants an infinite "sum" (or sequence) one may just take a formal serie. That is a ring, hence also an abelian group.
93090
I am interested in the result above, and would like to study a textbook that delves into it deeper.  However the standard algebraic topology textbooks (e.g. Hatcher/Munkres) and standard algebra books like Hungerford do not contain this result.  Is there any suitable book that I can consult? Thanks!
33488
I was wondering if there is a reason why in the definition of a topologic space, the reunion of any collection of open sets is also an open set but the intersection has to be finite to be a set.  Is there because of the definition of an open set itself ? I have always seen it as a definition, but now I want to know why.  Thanks for your help.