Suppose that $C$ is the cantor set and let $\sim$ be an equivalence relation on $C$. Are there necessary and sufficient conditions on $\sim$ to guarantee that the quotient $C/\sim$ is Hausdorff?  What I know:   If $q: X \to Y$ is an arbitrary quotient map then $Y$ is $T_1$ if and only if the fibres of $q$ are closed. So the equivalence classes of $\sim$ must be closed in $C$. Since $C$ is compact and normal, the results listed in   imply that if the quotient map $q: C \to C/\sim$ is closed then $C/\sim$ will be Hausdorff. What conditions can be put on $\sim$ to guarantee that the quotient map is closed?
5456
$X$ is a Hausdorff space and $\sim$ is an equivalence relation.  If the quotient map is open, then $X/{\sim}$ is a Hausdorff space if and only if $\sim$ is a closed subset of the product space $X \times X$.  Necessity is obvious, but I don't know how to prove the other side. That is, $\sim$ is a closed subset of the product space $X \times X$ $\Rightarrow$ $X/{\sim}$ is a Hausdorff space. Any advices and comments will be appreciated.
103874
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?
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I'm thinking about Hausdorff spaces, and how mappings to Hausdorff spaces behave. Suppose I have an arbitrary (continuous) function $f:X \longrightarrow Y$, where $Y$ is a Hausdorff space (I think it is irrelevant for my question whether $X$ is Hausdorff or not, so I just consider it to be a topological space - if this is incorrect, please correct me!).  Can we say that the graph $$G_f=\{(x,f(x)) \in X \times Y\ : x \in X \}$$ is a closed subset of $X \times Y$? It seems quite obvious that it is the case, but I cannot see how to prove it. If anyone can offer a proof I'd be very interested. Regards.  EDIT 1  In response to Hennning Makholm:   I wasn't really aware of any variation in 'definition'; I guess I'm considering closed sets to be those with an open complement (though naturally this definition gives rise to other definitions, such as the subset equalling its closure etc.). For continuity of such a map, I would normally consider continuity to mean that $f^{-1}(V)$ is closed in $X$ whenever $V$ is closed in $Y$, though again definitions involving convergence of sequences and the notion that $f$ is continuous iff $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$ are also known to me.
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Let $f: X \to Y$ be continuous and proper (a map is proper iff the preimage of a compact set is compact). Furthermore, assume that $Y$ is locally compact and Hausdorff (there are various ways of defining local compactness in Hausdorff spaces, but let's say this means each point $y \in Y$ has a local basis of compact neighborhoods).  Prove that $f$ is a closed map.  I know that this proof cannot require much more than a basic topological argument. But there's just something that I'm missing.   We can start with $C \subseteq X$ closed, and then try to show that $Y \setminus F(C)$ is open (for each $q \in Y \setminus F(C)$, we would want to find an open set $V_q$ with $q \in V_q \subseteq Y \setminus F(C)$).   Hints or solutions are greatly appreciated.
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This is an old exam question that I don't have a solution to:  Let $X$, a compact Hausdorff (T2) space, and let $\phi$ a family of closed, non-empty, and connected subsets of $X$, such that for every $A, B \in \phi$, $A \subset B$ or $B \subset A$.  Prove that $Y:= \cap \{A: A \in \phi\}$ is connected.   I tried to solve this question with a friend, and this is what we came up with:   Obviously, $X$ is normal space (T4). We tried to see what happens if $Y = U_1 \cup U_2$, disjoint and open sets (and?) We tried to use nets and maybe see if we can create a net that converges to $x \neq y$ (and contradict X being Hausdorff space). We tried to work with continuous functions, but this idea didn't lead us anywhere either.   I feel like there is a simple observation that we are missing. Any ideas?  Thanks!
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Suppose $X$ a metric space and $\sim$ an equivalence relation on $X$. Is the space $X/\mathord{\sim}$ metrizable? I think that the answer is no, but I could not arrive at a counterexample.
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There is this example at the Wikipedia article on Quotient spaces (QS):     Consider the set $X = \mathbb{R}$ of all real numbers with the ordinary topology, and write $x \sim y$ if and only if $x−y$ is an integer. Then the quotient space $X/\sim$ is homeomorphic to the unit circle $S^1$ via the homeomorphism which sends the equivalence class of $x$ to $\exp(2πix)$.   I understand relations, equivalence relation and equivalence class but quotient space is still too abstract for me. This seems like a simple enough example to begin with.  I understand (sort of) the definition but I can't visualize. And by this example and others, there is a lot of visualizing going on here! torus, circles etc.
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Let $(B_i)_{i\in I}$ be an indexed family of closed, connected sets in a compact space X. Suppose $I$ is ordered, sucht that $i &lt; j \implies B_i \supset B_j$.  Is $B = \bigcap_i B_i$ necessarily connected?  I can prove it, if I assume $X$ to be Hausdorff as well: If $B$ is not connected, then there are two disjoint, closed, nonempty sets $C$, $D$ in $B$, such that $C \cup D = B$. Now these sets are also closed in $X$, hence by normality there exist open disjoint neighborhoods $U$, $V$ of $C$ and $D$, respectively.  Then for all $i$: $B_i \cap U^c \cap V^c \ne \emptyset$, since $B$ is contained in $B_i$ and $B_i$ is connected. Thus we must also have  $$ B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset $$  by compactness and the fact that the $B_i$ satisfy the finite intersection property. This is a contradiction to the choice of $U$ and $V$.  I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated!  Thanks,  S. L.
87173
Let $X$ be infinite set and $a \in X $. Let $$\mathscr{F} = \{ A \subset X: |A| \; \; finite \; \;, a \notin A \}$$  Then $\mathscr{F}$ is the set all of all closed sets for some topology $\mathcal{T} $ on $X$. Also, any two continuous maps $f,g: (X, \mathscr{T}) \to Y$ ( $Y$ is metric space) have the property that if $f(b) = g(b)$ for all $b \neq a $, then $f(a) = g(a)$.  In my notes my professor posted this theorem without proof, so I was wondering what is best way to proof this theorem.
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Let $C^\prime$ be the Cantor set and let $C = C^\prime \times \{0\}$ (viewed as a subset of $\mathbb{R}^2$). For $c \in C$, let $L(c)$ denote the half-closed line segment connecting $(c,0)$ to $(\tfrac{1}{2}, \tfrac{1}{2})$ (including $(c,0)$ but excluding $(\tfrac{1}{2}, \tfrac{1}{2})$).  If $c \in C$ is $(0,0)$, $(1,0)$, or an endpoint of an interval deleted in the Cantor set, let $X_{c} = \{ (x,y) \in L(c) : y \in \mathbb{Q} \}$. For all other $c \in C$, let $X_{c} = \{ (x,y) \in L(c) : y \notin \mathbb{Q} \}$.  Cantor's Teepee is the set $\bigcup_{c \in C} X_{c}$ equipped with the subspace topology inherited from the standard topology on $\mathbb{R}^2$.  I've found several references to the fact that Cantor's Teepee is totally disconnected, but I cannot quite prove it.     It was brought to my attention in the comments that I had taken the wrong definition for totally disconnected when first working on this problem, as demonstrated in the following example. The property I had been attempting to prove was , which lead to the quandary I was experiencing below.     In particular, I cannot seem to separate two points that belong to the   same $X_c$. For concreteness:           What are two separated open sets $A$ and $B$ such that               $A$ and $B$ witness the disconnectedness of Cantor's Teepee,     $(0,0) \in A$, and     $(\tfrac{1}{4}, \tfrac{1}{4}) \in B$?              Notice $(0,0), (\tfrac{1}{4}, \tfrac{1}{4}) \in X_0$. Presumably the   method that works for this example can be easily modified to work for   any pair of points belonging to the same $X_c$.
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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 $X$ be a compact Hausdorff space. Show that the cone in $X$ is homeomorphic to the compactification of $X \times [0,1)$. If $A$ is closed in $X$, show that $X/A$ is homeomorphic to the compactification of $X\setminus A$.  I don't know even how to begin to solve this question, it's seems so hard, anyone could help me, please.
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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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In the post  , it is argued that if $X$ is second countable and locally compact Hausdorff, then $C_{0}(X)$ is separable. Is it also true that $C(X)$ is separable under the same hypotheses?
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Suppose that $X$ is a topological space equipped with a delta-complex structure. Prove that $X$ is Hausdorff.  I can actually "see" why it should be Hausdorff after the hint from Hatcher asks to "think of a delta-complex X as a quotient space of a collection of disjoint simplices".  I would appreciate if somebody can write out a full formal argument showing this proof. I'm trying to start from the definition of Hausdorff but I don't know how to carry on.
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I am reading a proof that the cantor set is uncountable and I don't understand it. Hopefully someone can help me.  Let $C$ be the Cantor set and $x\in C$.  Then there exists unique $x_k\in \{0,2\}$ such that $x=\sum_{k\in \mathbb N}\frac{x_k}{3^k}$. Conversely every $x$ with this representation lies in C.  If $C$ would be countable then there would exists a injective map $f:C\rightarrow \mathbb N$.  Let $f^{-1}(i)=\sum_{k\in \mathbb N}\frac{x_{ik}}{3^k}$ or $f^{-1}(i)=\emptyset$.  Set $z_k=2-x_{i_k}$, then z=$\sum_{k\in \mathbb N}\frac{z_k}{3^k}$  Then $z$ is in the cantor set but z isn't in the pre-image of $f$.  Questions:   Shouldn't it be $z_k=2-x_k$? I understand that z lies in C but why it is not in the pre image of f?   Thanks in advance
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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.