We can define the cofinite topology as the following $\mathcal{T}_{cofinite} = \{U \subseteq X| U = \emptyset, X-U  \mbox{is finite}\}$. Let $\mathcal{U}= \{U_{\alpha}\}_{\alpha \in I}$ be an open cover for $\mathbb{R}$. But where can we take this argument. We want to find a finite sub cover for $\mathbb{R}$ and then we are done. We can also included that since $\mathcal{U}= \{U_{\alpha}\}_{\alpha \in I}$ is an open cover for $\mathbb{R}$ that $\mathbb{R} = \cup_{\alpha \in I} U_{\alpha}$.
97390
While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)?   What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology?
72716
Let $\mathfrak{C}$ be an open cover of a topological compact space $X$, and let $\mathfrak{B}\subseteq \mathfrak{C}$ be its finite subcover. Then every subset of $X$ also has the same cover and subcover! Shouldn't every such subset also be compact then?  Motivation: I read somewhere that every closed subset of a compact space is compact. I wonder why it shouldn't be the case for every subset of $X$, rather than just the closed ones.
30251
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.
3999
Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.  What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$?
119802
How to prove for every open cover of the real numbers $\mathbb{R}$ there is a countable subcover? Without using more sophisticated results from topology, assuming only a real analysis background.  I've found a  using second-countable space characterization, but since i never studied general topology before, it's hard to associate a countable base on the real line. My intuition says to transform the open cover into disjoint open subsets, but how to achieve that?
83766
I´m studying Cech cohomology through Hartshorne´s algebraic geometry and EGA, and I´m stuck in two exercises, I wonder if you can help me. The first one is the exercise III.4.4, we want to prove that $$\varinjlim_{\mathfrak{U}}\check{H}^{1}(\mathfrak{U},\mathcal{F})=H^{1}(X,\mathcal{F})$$ where $\mathfrak{U}$ is a covering of $X$. (I attach a photo of the exercise)   I have just proved $a)$ and $b)$.  I´m also stuck with the last exercise of this section: Let $X$ be a topological spaces, $\mathcal{F}$ a sheaf of abelian groups, and $\mathfrak{U}=(U_{i})$ an open cover. Assume for any finite intersection $V=U_{i_{0}}\cap\cdots\cap U_{i_{p}}$ of open sets of the covering, and for any $k&gt;0$, that $H^{k}(V,\mathcal{F}_{\mid V})=0$. Then prove that for all $p\geq 0$, the natural maps  $$\check{H}^{p}(\mathfrak{U},\mathcal{F})\rightarrow H^{p}(X,\mathcal{F})$$ of the last exercise (III. 4.4) are isomorphisms.  Thank you for your time.
65126
Prove or find counterexamples.  Let $X$ be an infinite set and $T$ be a topology on $X$. If $T$ contains every infinite subset of $X$, then $T$ is the discrete topology.
93827
I am new to compact sets, and I had some hard time trying to solve this:  Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.  I know that I may consider the open cover $\{\{x\},x \in X\}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.  Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?  Any help is appreciated.
77548
I'm trying to prove the following statement:     Let $X$ be a topological space with a countable open cover $C= \{U_\alpha : \alpha \in A\}$. If each $U_\alpha$ is second-countable then $X$ is second-countable.   However I'm stuck on how to proceed, so any hints on how to tackle it would be highly appreciated.
109363
I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.
110236
Let S be an infinite subset of H  Assume that no point of H were a limit point of S.  Since H is compact, there exist a collection of open sets V={Vq: q∈H }  Vq∩S is finite (Is this true? If yes why?)  Since S is assumed to be infinite  ⋃Vq ⊃ S cannot cover S  Since S⊂H  ⋃Vq ⊃ H (cannot cover H)  This contradict compactness of H.  Hence every infinite subset of H has a limit point (Why do we conclude this?)
45780
question is regarding the same problem. I wish to present my proposed solution and get feedback on my argument, and as such, I claim that it is not a duplicate. (In particular, the other asker actually knew what he was doing, so this question will probably be sufficiently different. It also includes the discussion about the box topology, which the original question lacked.) We are given that $\{x_n\}$ is a sequence of points in the product space $\prod X_\alpha$. We are to show that this sequence converges to $x$ if and only if $\{\pi_\alpha(x_n)\}$ converges to $\pi_\alpha(x)$. Moreover, we are to determine if this holds if we equip $X = \prod X_\alpha$ with the box topology. We first show $\implies$. Suppose, for the sake of contradiction, that $\{\pi_\alpha(x_n)\}$ does not converge to $\pi_\alpha(x)$ for a fixed $\alpha$. Then there exists an open set $U_\alpha$ containing $\pi_\alpha(x)$ such that $X_\alpha - U_\alpha$ contains infinitely many $\pi_\alpha(x_n)$. Since every projection in the product topology is continuous, $\pi^{-1}(U_\alpha)$ is open in $X$. This contradicts the assumption that $\{x_n\} \to x$, since we now have an open set of $X$ where one of the $\{x_{n_i}\}$ clearly do not converge, where $\{x_n\} = (\{x_{n_1}\},\{x_{n_2}\},\cdots)$ We proceed to show $\impliedby$. We are given that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$ for every $\alpha.$ Let $\prod U_\alpha$ be an open set in $X$ containing $x$. By the definition of the product topology, we know that $U_{\alpha} \ne X_\alpha$ for only finitely many $\alpha_1, \alpha_2,\cdots,\alpha_k$. Now, since we know that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$, it follows by the definition of convergence that there exist numbers $N_{\alpha_1},N_{\alpha_2},\cdots,N_{\alpha_k}$ such that $\pi_{\alpha_i}(x_n) \in U_{\alpha_i}$ for $i \in \{1,\cdots,k\}$ and $n \ge N_{\alpha_i}$. Letting $N = \max\{N_{\alpha_1},\cdots,N_{\alpha_k}\}$, we see that $x_n \in \prod U_\alpha$ for $n \ge N$. (Recall that for $\alpha$ not in the finite collection, it equals the whole space $X_\alpha$, where inclusion is trivial.) We conclude that $\{x_n\} \to x$ Lastly, we consider $X$ with the box topology. In $\implies$ we did not use any particular properties of the product topology, aside from the continuity of projections, which also holds for the box topology. As such, I claim that $\implies$ still holds. However, I claim that $\impliedby$ does not hold. I ask for an example that shows this, since I am not able to find any argument I'm happy with. (&quot;Well, our proof used properties of the product topology, so...&quot; is only acceptable for my private notes.)
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 $i$?
98862
I'm totally stuck with the following question, I even don't know how to start:  For each natural number $n$ we consider a space $X_n$ that is obtained by removing $n$ distinct points from $\mathbb{R}^2$. We consider the 1-point compactification $X_n^+$ and we denote by $\infty_n\in X_n^+$ the point at infinity (so that $X_n^+=X_n\cup \{\infty_n\}$). Show that   $X_n^+$ can be embedded in $\mathbb{R}^3$. I do not have to write down explicit formulas for the embedding, but I have to explain my reasoning using pictures and mention what result(s) I use in order to reach my final conclusion If $X_n$ and $X_m$ are homeomorphic, then $n=m$.    Both questions I find very hard to answer. I really hope someone can help me!
12509
Is the cofinite topology on a countable set first countable? I know it is not first countable on R but I don't know the case of a countable set.
25276
The question is:     Show that if $A⊆\Bbb R^n$ is finite and $n≥2$, then the set $\Bbb R^n\backslash A$ is connected in $(\Bbb R^n,τ_d)$ where $τ_d$ is the topology induced by the Euclidean distance in $\Bbb R^n$.   My attempt:  Denote $A^C=\Bbb R^n\backslash A$ and suppose $A=\{a_1,…,a_n\}$. Assume by contradiction that $A^C$ is not connected, then there exists $O_1,O_2∈τ_{A^C }$ st. $A^C=O_1⋃O_2,O_1⋂O_2=∅$. Here $\tau_{A^C}$ is a subspace topology.  Intuitively, part of $A$, denoted by $A_1$ is "enclosed" in $O_1$ and the rest, denoted by $A_2$ are "enclosed" in $O_2$. Then $G_1=O_1\bigcup A_1$ is an open set in $\tau_d$, and $G_2=O_2\bigcup A_2$ is also an open set in $\tau_d$, and $\Bbb R^n=G_1\bigcup G_2$. However, $G_1, G_2$are disjoint, contradicting with the fact that $\Bbb R^n$ is connected.  However I got stuck when trying to implement this idea. Can anyone provide some help? Thank you!
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 $A \supset B$.