Let us consider $(X,\tau)$ a Hausdorff topological space y let $\infty$ an element such that $\infty \notin X$. Then $(\hat X, \hat \tau)$ is topological space, where $\hat X= X \cup \lbrace \infty \rbrace$ and $\hat \tau=\tau \cup \lbrace \hat X-C :$ C is a compact subset of $(X, \tau) \rbrace$. Well. Now I have to show that:  Be $(X_1,\tau_1)$ and $(X_2,\tau_2)$ Haussdorf spaces, locally compact and are not compact. If $(X_1,\tau_1)$ and $(X_2,\tau_2)$ are homeomorphic then $(\hat X_1,\hat \tau_1)$ and $(\hat X_2,\tau_2)$ are homeomorphic. My attempt. If $f$ is a homeomorphism between $X_1$ and $X_2$ then we define $\hat f : (\hat X_1,\hat \tau_1)\rightarrow (\hat X_2, \hat \tau_2)$ by $\hat f(x)=f(x),\forall x \in X_1$ and $\hat f(\infty_1)=\infty_2$. Clearly this function is bijective but I can't show that $\hat f$ is continuos and an open (or closed) map.  Any hint can be useful. Thanks
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Say $X^{*}$ represents the one-point compactification of a space $X$ that is locally compact, non-compact, and Hausdorff. Also suppose that $Y$ is compact Hausdorff and $Y-\{p\}$, for a single point $p$, is homeomorphic to $X$. So then would $Y$ be homeomorphic to $X^{*}$? I think it is, but I'm not sure how it can be shown.    My attempt so far: Correct me if I'm wrong, but I started with inclusions $i: X \rightarrow X^{*}$ and $j: Y-\{p\} \rightarrow Y$, along with the homeomorphism $h: X \rightarrow Y-\{p\}$. So we need to find a homeomorphism $g: X^{*} \rightarrow Y$. So I tried using a commutative diagram and so far, I've concluded that $g$ (if it is to exist) would have to be continuous since the continuous image of a compact space is compact (I think this follows from $X^{*}$ being compact), but then one of my inclusions, i.e. $i$, can't possibly be continuous, can it (since $X$ is not compact)? How can this be fixed?
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Theorem A map $\phi$ maps a topological space $X$ to another $Y$, where $X$ is Hausdorff, $Y$ is compact and the graph of $\phi$ is closed. Then $\phi$ is continuous.   Is it reallly necessary to include the condition that $X$ is Hausdorff?  Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer.            A proof:     For any closed subspace $C$ of $Y$, the pre-image $D$ ought to be closed. For any element $a$ in the complement of $D$, we can use the compactness to show that there is a finite number of open sets in $Y$ such that the union of them covers the C, and then the corresponding open sets in $X$ is an open neighborhood of $a$ which has an empty intersection with $D$; so $D$ is closed.     P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that $(a,c)$ is not in the graph for any $c$ in $C$.
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Let $\mathbb{H}^n = \{(x^1, \dots , x^n) \in \mathbb{R}^n : x^n \ge 0\}$, and $\overline{\mathbb{R}}^n_+ = \{(x^1, \dots , x^n) \in \mathbb{R}^n : x^1 \ge 0, \dots, x^n \ge 0\}$. Endow each set with the subspace topology it inherits from $\mathbb{R}^n$.  Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $\mathbb{H}^n$ and $\overline{\mathbb{R}}^n_+$ are homeomorphic.   However, it seems to me the spaces are not homeomorphic, as shown by the following argument.  Suppose $f:  \overline{\mathbb{R}}^n_+ \to \mathbb{H}^n$ is a homeomorphism. Set $f(0, \dots , 0) = (a^1, \dots, a^n)$. Then   $$A = f(\overline{\mathbb{R}}^n_+ \setminus \{(0, \dots, 0)\}) = f((\infty,0) \times \cdots \times (\infty,0)) = \\  ((\infty, a^1) \cup (a^1, - \infty)) \times \cdots \times ((\infty, a^n) \cup (a^n, 0]),$$ where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.  The set $A$ is connected as it is the continuous image of the connected set $(\infty, 0) \times (\infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets  $(\infty, a^1) \times \cdots \times (\infty, a^n)$ and $ (a^1, - \infty) \times \cdots \times (a^n, 0]$ (or $(\infty, a^1) \times \cdots \times (\infty, 0)$ and $(a^1, - \infty) \times \cdots \times (\infty, 0)$ in case $a^n = 0$).   Is there a  mistake in my argument or something I am missing? Any comments are greatly appreciated.
11551
I am trying to solve the following problem.     Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous.   Show that there exists a non-empty set $A\subset X$ such that   $f(A)=A$.   There is a hint to define $A=\cap_{n\ge 0} A_n$ with $A_{n+1}=f(A_n)$ and $A_0=X$.  Then $x\in A_{n+1}$ implies that $f^{-1}(x)\cap A_n\ne\emptyset$. Right? So if $f^{-1}(x)\cap A_n=\emptyset$ then $x\notin A_{n+1}$. Is this correct?  First I need to show that for $A$ defined in the hint it holds $f(A)=A$. For this to happen it is enough to have $f(A)=\cap_{n\ge 0} f(A_n)$ Because then $f(A)=\cap_{n\ge 1} A_n=\cap_{n\ge 1} A_n\cap X=A$. But got stuck proving the equality.  What I did for that is to assume that there exists $x\in A$. This means that $\exists y_n\in A_n$ such that $f(y_n)=x$ for all $n$. Is this enough to deduct that $f(x)\in A$?  Then I need to show that $A$ is not empty. For this I think it is enough to show that $A_{n+1}\subset A_n$, because $A_n$ are compact and then their intersection is not empty. However I am stuck there too. I tried by contradiction assuming $\exists x\notin A_n$ with $f(x)\in A_{n}$, but I cannot work from there.  Initially I thought of something else. Choose $x_n\in A_n$ and create the sequence $\{x_n\}$. This has a convergent subsequence $\{x_{n_k}\}$. Then I create a sequence $\{x'_m\}$ by defining $x'_m=x_{n_k}$ with $n_k=\max\{j\in\{n_k\}|j\le m\}$. So basically i repeat the terms of the subsequence to get a convergent sequence. However this again requires that $A_{n+1}\subset A_n$ so I have $x'_m\in A_m$. Then I think that the limit $x'_m\to x$ is in $A$ but I was not able to show that also.  Thanks in advance.
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...........
56849
I was asked the following question: if two topological spaces $X, Y$ are such that there exist a function $f:X\rightarrow Y$ continuos and bijective and a function $g:Y\rightarrow X$ continuous and bijective, then are the two spaces homeomorphic?  I would say that the answer is No, but I didn't find any counterexample so far.
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I was given this question on my midterm. Currently I am studying for finals and am still unsure how to properly solve this question.  Let X and Y be two sets and f be a map from X to Y be a bijection. Prove that, when we consider X and Y with their respective T1-topologies (cofinite topology), this map is a homeomorphism.   I know that I must show that f is continuous in order to show X and Y are homeomorphic. On the midterm I tried showing that open sets in Y have preimage that are open in X and closed sets in Y have preimage that are closed in X. However I'm unsure how to approach the sets in X,Y that are neither open nor closed.  So my question is, would it be easier to show continuity by first showing local continuity at every point in X and using that to prove the continuity of f? Or is this the wrong way to approach this problem?
123581
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
74996
I wish to show that any metrizable space $(X,\mathcal{T})$ is Hausdorff  Proof attempt:  Let $d$ be the metric that generates the topology on $X$. Pick two points $x,y \in X$, we wish to produce two disjoint open sets that separates $x,y$.  Let $B_\epsilon(x)$ and $B_\delta(y)$ be two metric balls containing $x,y$ respectively. Suppose that  $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$       Stuck Here: Hmm...How should I adjust $\epsilon, \delta$ so that these   balls are separated?   Idea: Reduce $\epsilon$ by half. If they are still intersecting...reduce $\delta$ by half. Continue ad infinitum         Is there a more satisfying solution i.e. closed form expression for   reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!
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Let be X a topological space and let be $Z,Y\subseteq X$ such that $Z\subseteq Y$: so if Z is compact in Y then Z is compact in X.  To prove the assertion let be U an open cover of Z in X and we consider the open cover $\mathcal{U}'=\{U\cap Y:U\in\mathcal{U}\}$ of $Z$ in $Y$ -indeed $Z\subseteq\bigcup\mathcal{U}=\bigcup_{U\in\mathcal{U}}U\rightarrow Z=Z\cap Y\subseteq(\bigcup_{U\in\mathcal{U}}U)\cap Y=\bigcup_{U\in\mathcal{U}}(U\cap Y)=\bigcup\mathcal{U}'$- thus there exist $\mathcal{U}''\subseteq\mathcal{U}'$ such that $|\mathcal{U}''|&lt;ℵ_0$ and $Z\subseteq\bigcup\mathcal{U}''$ and so $\mathcal{U'''}=\{U\in\mathcal{U}:U\cap Y\in\mathcal{U}''\}$ is an open and finite cover of Z in X -indeed $|\mathcal{U}'''|=|\mathcal{U}''|&lt;\aleph_0$ and $Z\subseteq\bigcup\mathcal{U}''\subseteq\bigcup\mathcal{U}'''$- and so this prove that $Z$ is compact in $X$.  It is my proof correct? if not, how prove the assertion? could be that if is false?  Could someone help me, please?
217
Suppose that $\tau_1$ and $\tau_2$ are two topologies on a set $X$ with the property that $K\subset X$ is compact with respect to $\tau_1$ if and only if $K$ is compact with respect to $\tau_2$. Then is this enough information to determine whether $\tau_1 = \tau_2$?  If not (which is suspect to be the case) is there a nice counterexample, and what is the minimum amount of extra information required for $\tau_1 = \tau_2$?  I feel like there would be an issue regarding 'points at infinity.'
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This question is from Wayne Patty Exercise 4.2  Let $T$ and $U$ are Topologies on set X. Suppose $ (X,T) $ and $(X,U)$ are compact Haurdsorff Spaces. Prove that $T=U$ or $T$ is not a subset of $U$ or $U$ is not a subset of T.  I thought of proving it by this way: Assuming that if $T$ is a subset of $U$ or $U$ is a subset of $ T$ then $T=U$ must be true. But I was unable to think on how to use compactness or hausdorff property and was unable to proceed. Well, also this question was asked before on this website. But I am unable to understand how the solution will work despite trying to understand it for a fair amount of time as it was just outlined. So, I am looking for a detailed solution.
109732
A topological space is locally compact then there is an open base at each point has all of its set with compact closure, how can i prove it?  It seems very meaningful when the topological space is metrizable and the topology is the topology induced by the metric, but i could not prove it for general topological spaces.  I am not familiarized with Hausdorff spaces.  What i tried for now is starting from the definition of locally compact as the topological space whose ever point has a neighbourhood with compact closure. I picked (for each point) this neighbourhood, and created a class with all open sets contained in that neighbourhood, then i thought that this class may be the open base at the analized point whose all sets have compact closure, but i could not prove that this class is an open base at point, have someone an idea? Or other path to prove it, or yet is this claiming false?
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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?
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Let $X$ be a Hausdorff space and let $C_0 \supset C_1 \supset ...$ be a decreasing sequence of compact connected subsets of $X$.  Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected.  I have the compact part.  Just don't know how to show it is connected.
8909
Be $S^1$ the unit circle in the plane, that is, $S^1=  \lbrace (x,y)  : x^2+y^2=1 \rbrace$ with the subspace topology. Show that $S^1 - \lbrace (1,0)\rbrace$ is homeomorphic to the open interval $(0,1)$.  I have an idea. Using this theorem:  Let $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ be a homeomorphism. Let $a \in X$, so that $X - \lbrace a \rbrace$ is a subspace of $X$ and has induced topology $\tau_3$. Also, $Y-\lbrace f(a)\rbrace$ is subspace of $Y$ and has induced topology $\tau_4$. Then $(X - \lbrace a \rbrace, \tau_3)$ is homeomorphic to $(Y-\lbrace f(a)\rbrace, \tau_4)$. (Morris, "Topology without tears", remark 4.3.6)  My idea is to make $X=S^1$ and $Y=(0,1) \cup ?$ so, with the respective topologies using the theorem, $S^1-\lbrace (1,0)\rbrace$ is homeomorphic to $(0,1)$.  But I'm stuck in the part of defining $Y$ and find the homeomorphism $f$ between $X$ and $Y$.  PD. Sorry about my English, but it is not my native language.
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Let $X$ be a topological space and let $\left\{ Y_{i}\right\} _{i=1}^{\infty}$  be a descending chain of closed connected subsets of $X$. I know from reading elsewhere that ${\displaystyle \bigcap_{i=1}^{\infty}Y_{i}}$ is not necessarily a connected subspace of $X$   but I have no counter example and I haven't managed to come up with one.  There is a counter example  to the same question while also assuming $X$ is compact. However, it uses the quotient topology which I haven't studied about so I would prefer a different counterexample that does not use the quotient topology.  Help would be appreciated!