I have the following question on a worksheet: Consider the finite complement topology τf on N. Does this topology have a countable base? If so give one such base and if not prove your claim.  Been trying to solve this for several hours but I have got nowhere. I think that there is no countable basis because it seems like to construct a basis you would need to take the entire topology, but I have no idea if that's right or how to show that.   Thanks
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I am working on the problems from the textbook "Topology without tears". I am stuck with problem number $4$ in Exercise $2.2$. Could anyone suggest some hints on how to proceed? The question goes as follows.  A topological space $(X,\tau)$ is said to satisfy the second axiom of countability or to be second countable if there exists a basis $\mathcal{B}$ for $\tau$, where $\mathcal{B}$ consists of only a countable number of sets.   Prove that the discrete topology on an uncountable set does not satisfy the second axiom of countability. Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfy the second axiom of countability?   For the first problem, I tried to argue by contradiction assuming that the basis for the topology is countable. But I did-not know what I need to look at to prove the contradiction. For the second one, I do not know where to start. I have not thought deeply on the second one though.  Thanks, Adhvaitha
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Let $\mathbb{Q}$ be the set of rationals, as relative topology to $\mathbb{R}$.   How to show that $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$?  More general, if $T$ is a countable dense space(no isolated poiont) with every singleton closed, how to show that $T^2$ is homeomophic to $T$.
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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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It is just an idea (might be wrong) but, i think that if a Hausdorff space, say $X$, contains too many elements, then a countable subset cannot be dense in it.  Does there exist a cardinality that any space with that (or bigger) cardinality cannot have a countable dense subset?  Thanks.
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Does someone know how to prove this? Let S and T be non-empty subsets of a topological space (X, τ) with S ⊆ T. (iii) Hence show that if S is dense in X, then T is dense in X. (iv) Using (iii) show that R has an uncountable number of distinct dense subsets. I just wanna prove the fourth item using the result of the third. I tried to do it by contradiction, but i don't know what exactly i have to do!
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Is the finite Cartesian Product of Countable sets countable?  If possible, could you give a bijection that would work for any example?  Any help would be appreciated.
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I am trying to prove that in $\mathbb{Z}$ with co-finite topology the only path-connected components are the singletons.  (I reckon that) showing that      "if a function $\gamma : [0,1] \to \mathbb{Z}$, where $\mathbb{Z}$ has   co-finite topology, is continuous then it is constant"   should do the trick.  However I am not sure this is true, let alone if this is a good approach to the problem. Any thoughts about it?  Edit: I was thinking: suppose $x,y \in \mathbb{Z}$ and $\gamma : [0,1] \to \mathbb{Z}$, where $\mathbb{Z}$ has co-finite topology. Further suppose $x \neq y$, then $f^{1}(\mathbb{Z}\setminus \{x\})=(0,1]$ which is not open in $[0,1]$, contradicting continuity. Hence $x=y$. Does it seem right?  Edit 2:  Forget the (stupid) edit above!
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Assuming the following: let $X$ be a set with two elements, $X$ = {$a,b$}.  what are all the possible topologies on $X$ ?  The answer I've come up with is: $$\tau_{T}=\{ \emptyset, X\} $$ $$ \tau_{a}=\{\emptyset,\{a\},X\} $$ $$\tau_{b}=\{\emptyset,\{b\},X \} $$ $$ \tau_{D}=\{\emptyset,\{a\},\{b\},X\}$$  My question is, are these correct? and are there any more?  thanks in advance
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Let $X$ be a connected, locally compact space and let $K$ be a compact subset of $X$. Is the number of connected components of the complement $X\setminus K$ finite? If it is not, are there some hypotheses over $X$ for finiteness of connected components?
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What is wrong with this reasoning: The union (finite or infinite) of countable sets is countable. The set of all binary sequences is the infinite union of the sets $S_n$ of all the binary sequences of length n, which are finite, hence countable.
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Let $X$ be any uncountable set with the cofinite topology. Is this space first countable?  I don't think so because it seems that there must be an uncountable number of neighborhoods for each $ x \in X$. But I am not sure if this is true.
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I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
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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.
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I have the set $$X = \{1, 2, 3\}$$  and I want to figure out how many different topologies I can get from the set $X$  so what I have done is assumed that the empty set and the whole set are in $T$  and that any union of elements will satisfy, $X \in T$, as well as any intersection of elements.  therefore all possibilities will be considered topologies.  so after I assume this, my answer I acheived was a simple $3!*3! + \emptyset = 37$.  Leading me to the fact that I believe there are 37 different topologies possibel from this set $X$.  Any help, disccrestions, criticims, or corrections would be great, i only have 75% confidence in this result.   thanks in advance
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I am having a difficulty setting up the proof of the fact that two basis of a vector space have the same cardinality for the infinite-dimentional case. In particular, let $V$ be a vector space over a field $K$ and let $\left\{v_i\right\}_{i \in I}$ be a basis where $I$ is infinite countable. Let $\left\{u_j\right\}_{j \in J}$ be another basis. Then $J$ must be infinite countable as well. Any ideas on how to approach the proof?
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Is $(\mathbb{R},\tau_{co})$ compact where $\tau_{co}$ is the cocountable topology on $\mathbb{R}$?  I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says:     $\mathbb{N}-\{n\}$ is infinite countable for all $n\in\mathbb{N}$.   Then $V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}$ for all $n\in\mathbb{N}$.      Then $U=\{ V_n : n\in\mathbb{N}\}$ is a open covering of $(\mathbb{R},\tau_{co})$.      $U$ does not contain a countable subcover since $V_n\cap\mathbb{N}=\{n\}$ for each $n\in\mathbb{N}$ and $\mathbb{N}$ is infinite.   How would you solve that problem?
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I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.  Proof: Suppse that $|B|&lt;2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $ \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.  What do you think?  Thank you, Shir