A covering space of a Hausdorff space is also Hausdorff. Conversely, a compact Hausdorff finite covering space has a Hausdorff base space. However, in general, a non Hausdorff space may have a Hausdorff covering space. Could someone give an example of the last? I can't find anyone. Thank you very much!
123235
Let $\overline{X} \rightarrow X$ be a covering space, where $\overline{X}$ is a Hausdorff space. Does $X$ have to be Hausdorff?  I assume there is a counterexample, but I couldn't find one yet.
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I know that every separable metric space has a countable base.  I was wondering if we can get a countable dense subset from every metric space that has a countable base.  Thank you very much!!
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Let $X$ and $Y$ be compact Hausdorff spaces. And let $A\subset X$ and $B\subset Y$ be closed subspaces. If $U$ is open in $X \times Y$ and $U$ contains $A \times B$, then there exists open subsets $V \subset X$ and $W \subset Y$ such that $A \times B \subset V \times W \subset U$.   I think this is an easier version of the generalized tube lemma since it assumes $X$ and $Y$ being Hausdorff, but I just can't seem to find such $V$ and $W$. I can find some open set in $X \times Y$ containing $A \times B$ but not able to show it is contained in $U$.
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Let $X$ be a topological space.  It is  that, if $X$ is compact, then every proper closed subset $Y \subset X$ is compact.  Out of curiosity, I would like to explore the converse of this statement:     Let $X$ be a topological space with the property that every proper closed subset is compact.  Is $X$ itself compact?   Of course, this question is only interesting when $X$ is infinite.  We can further restrict ourselves to spaces that have infinitely many open sets since every closed set, and indeed the space itself, would be compact otherwise.  If this result is false in general, what hypotheses can be added to make it so?  Should this not hold, I'm interested in finding concrete examples of when it does.  For instance, we can take any infinite set endowed with the , which works because every closed set is finite and thus compact, and the space itself would be compact.  Another example is an interval $[a, b] \subset \mathbb{R}$, wherein closed subsets of this are closed and bounded in $\mathbb{R}$; as such, they are compact per .    Update:  Zardo's answer reveals that such a topological space is indeed compact. For further discussion on the  and its relevance to compactness, see Munkres' : Chapter $3$, theorem $26.9$.
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Let $X$ be a normal Hausdorff space and let $C,D$ be two $F_{\sigma}$ subsets of $X$ such that $\overline{C} \cap D = \emptyset$ and $C \cap \overline{D} = \emptyset$.  Prove there exists disjoint open subsets $U,V$ such that $C \subset U$ and $D \subset V$.  No idea how to show this. Can you please help?
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I am looking for easy(!) counterexamples that the product of two quotient maps is not necessarily a quotient map and that the quotient space of a Hausdorff space is not necessarily Hausdorff.
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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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I know that this is separable, but that has not been covered by the text at this point. I believe my proof (below) to be correct, but not very rigorous. Assuming that it actually is correct, could someone please help me make it more formal? (If it's not correct, please help with the proof). Thanks.  Proof: Let $X$ be a compact metric space. Let $B_n = \{B(x,\frac1{n}) : x\in X\}$. be the collection of open balls centered at $x$. Let $\{x\}^{(n)}$ be the set of centers for some particular $n$. Then by compactness of $X$, there are also a finite number of centers for a particular $n$ (corresponding to each open ball).   [This is where I have trouble properly saying what's on my mind:]  Consider $B(x,r_0)$, where $r_0 = \frac{1}{n_0}$. When we decrease the magnitude of $r_0$ to $r_1$, we must still be able to cover $X$. Meaning some open ball(s) must be "formed" where $B(x,r_0)$ was once "covering". So by adding open balls, we also add to our collection of centers. This means at $B(x,r_0)$, there was originally a point $y \in B(x,r_0)$ s.t. $B(y,r') \subset B(x,r_0)$ for some $r'$. That is, $B(x,r_0)$ contained another point of $X$. Thus, by definition, $X$ contains a countable dense subset.
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Let $X$ be an infinite set. Show that the intersection of all Hausdorff topologies on $X$ is a topology, and find it (Hint: We are familiar with this topology).   The best candidate seems to be the co-finite topology on X, as it is a subset of each of the Hausdorff topologies, and it fits the hint. However, I am having trouble proving that for every subset of $X$ whose complement is infinite, there is a Hausdorff topology which does not include it.
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I was looking up the proof that a closed subset of a compact space is itself a compact space (with the induced topology). I came across this one:   I believe I have found one step that might be something of an inaccuracy. How do we know that the open sets that comprise the open cover of $C$ are open in the ambient space $T$?  An example of what I mean: Say $T=[0,2]$ with the standard topology inherited from $\mathbb{R}$. And $C\subset T$ is given by $C=[0,1]$. Then $C$ is closed in $T$. Now $C$ is an open cover of itself but $C$ is not open in $T$. According to the proof $C \cup (T\setminus C) = [0,1] \cup (1,2] $ is then an open cover of $T$, but this is clearly not the case.  All help would be highly appreciated.
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Show that the space of functions $f: \mathbb{N} \to \mathbb{R}$ does not have a countable basis.  I really don't know where to start with this one! Could anyone help me?  Thanks
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I am trying to find examples of maps between topological space which are local homeomorphism but not covering maps. Especially, how twisted has to be such a counterexample : can it be a local diffeomorphism between connected manifolds which is not a covering map ?   I found here() a nice proposition which state that a local homeo from a compact space to a connected Hausdorff space is a covering map.   I am interested in all type of counterexamples, from non-Hausdorff spaces to surfaces, to get a better picture of the differences between covering maps and local homeomorphisms.  Thank you !
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Let $X$ is locally compact Hausdorff space .If $C_{c}(X)$ is complet,then $X$ is compact (this is to be proved). I know that $C_{c}(X)$ is dense in $C_{0}(X)$. As $C_{c}(X)$ is complete implies that $C_{c}(X)=C_{0}(X)$. and I know only urysohn's lemma and Tiez's extension theorem on locally compact Hausdorff spaces and other basic things.
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Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous. Prove that there exists a non-empty closed set $C$ such that $f(C) = C$.    Proving by contradiction seems hard to me, since if $f(C)\neq C$ for all closed sets $C$, then we can have either $f(C)\subset C$, $f(C)\cap C\neq\varnothing$ or $f(C)\cap C = \varnothing$. The only thing I might have thought about is to set $C = f(X)$. Since $X$ is compact, $f(X)$ is compact and since $X$ is Hausdorff, $f(X)$ is closed. Then we have that $f(C)\subseteq C$ by definition. I was not able to prove that $C\subseteq f(C)$, namely that $f(X) \subseteq f(f(X))$. It is not obvious for me that in general it is even true, so maybe the choice $C = f(X)$ does not help.
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Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space,  Hausdorff?     I think it is true since if $a $ and $b $ are two distinct points, they can be separated two disjoint  open sets, since the main space is totally disconnected (see Theorem at &lt;  >).   Is this argument true?
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Suppose a $T_4$ space $X$ satisfies a property that every continuous map $f$ from $X$ to any Hausdorff space $Y$ is a closed map. Show that $X$ is compact. What I am thinking is to let $\mathcal{U}$ be an open cover of $X$ and find a way to choose finite open sets from $\mathcal{U}$ that can cover $X$ again. Do we need to define a suitable continuous map by ourselves and use the fact that it is a closed map to choose suitable open sets from $\mathcal{U}$? I have tried to define a continuous map $f \colon X \to Y$ as mapping different open sets from $\mathcal{U}$ to different points in $Y$. However, different open sets from $\mathcal{U}$ may intersect with each other and hence $f$ may not be well-defined. Apart form that, I have no idea about how to choose suitable $f$ and $Y$. Can someone help me please? Edit 1: I am trying to apply the Urysohn Lemma or the Tietz Extension Theorem due to the normality of $X$. But still have no idea. Edit 2: Finally found a way to prove without using compactification. I will post the proof when I’m free. Hint: Use Urysohn Lemma and Tychonoff Theorem.
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I know that the vector space of all real valued continuous functions on a compact Hausdorff space can be infinite dimensional.  When will it be finite dimensional?  And how will I identify that vector space with $\mathbb{R}^n$ for some $n$?