My attempt at specifying a homeomorphism is as follows.   Let $(X,\tau), (X\times[0,1]/\sim, \tau')$ and $(X\times[0,1),\tau'')$ be topological spaces. Let p be the equivalence class of $X\times\{1\}$ in $X\times[0,1]/\sim$. Let $\pi$ be the quotient map from $X \times [0,1]$ to $(X\times[0,1]/\sim, \tau')$.  Let $(\chi^+,\tau''^+)$ be the one-point compactification of $X\times[0,1)$, i.e. $\chi^+ := (X\times[0,1))\cup\{\infty\}$.   Let $f: (X\times[0,1]/\sim, \tau') \to (\chi^+,\tau''^+)$ be defined as: $$ f(x) = \begin{cases} x,  &amp; \text{if $x \in X\times[0,1)$} \\ \infty, &amp; \text{if $x =p$} \end{cases} $$    I claim that f is the homeomorphism that we are after.   That f is bijective is easy to see. Now I want to show that f is continuous.  Let $U \in \tau''^+$. We want to show that $f^{-1}(U)$ is open. If $\{\infty\} \notin U$, then $U$ is already in $\tau'$ and we are done.  Otherwise, let $U = ((X\times[0,1)) \setminus C) \cup \{\infty\}$, where $C \subset X\times[0,1)$ is a closed, compact set.   We have: $f^{-1}(U) = f^{-1}(((X\times[0,1))\setminus C) \cup \{p\} = ((X\times[0,1))\setminus C) \cup \{p\} $. And so $\pi^{-1}(f^{-1}(U)) = \pi^{-1}(((X\times[0,1))\setminus C) \cup \{p\}) = ((X\times[0,1))\setminus C) \cup X\times\{1\} = X\times[0,1]\setminus C$.    Since C is closed, we have that $X\times[0,1]\setminus C$ is open in $X\times[0,1]$. So $((X\times[0,1))\setminus C) \cup \{p\}$ is open in the quotient topology of $(X\times [0,1])/\sim$, i.e. $f^{-1}(U)$ is open in $(X\times [0,1])/\sim$. This proves the continuity of f.  Before proceeding with the continuity of $f^{-1}$, I'd like to ask for others' opinion on the part above. Also, just to be sure, in $X\times [0,1]$, are we considering $[0,1]$ as a subset of $\mathbb{R}$, i.e. does it has the subspace topology induced by the usual topology on $\mathbb{R}$? If not what is the topology on $[0,1]$?
124127
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.
92345
Let $f: X \to Y$ be a bijective continuous function.  If $X$ is compact, and $Y$ is Hausdorff, then $f:X \to Y$ is a homeomorphism.   My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively.  I want to know if my counter examples are correct:  Let $f: X \to Y$ be the identity function.    (1)  Let $X=[0,1]$ with the standard topology and let $Y=[0,1]$ with the indiscrete topology.  Then $f: X \to Y$ is continuous and bijective but not a homeomorphism.  Since $U=(1/3,1/2)$ is open in $X$, but $(f^{-1})^{-1}(U)=f(U)$ is not open in $Y$.  (2) Let $X=Y=\mathbb{R}$.  Let $Y$ have the standard topology and $X$ have the discrete topology.  Let $U=\{x\}$ since open sets are singletons.  $f(^{-1})^{-1}(x)=f(x)$ is closed in $Y$ since $Y$ has the standard topology.  Hence $f$ is not a homeomorphism.  This is the old question My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively.  I want to know if my counter examples are correct:  Let $f$ be the identity function.  Then $f: X \to Y$ be a bijective continuous function.  If $X=[0,1]$ with the standard topology and $Y=[0,1]$ with the indiscrete topology we have that $f:X \to Y$ is not a homeomorphism.  Let $f: X \to Y$ be a bijective continuous function, $X=(0,1)$, and $Y=[0,1]$ both with the standard topology.  It's not a homeomorphism because $f^{-1}([0,1])$ is not connected, but $[0,1]$ is connected.  The first shows why Hausdoff is necessary for $Y$ and the second shows compactness is necessary for $X$.  Are both of the examples correct?  I have no confidence in my first example because I'm having problems showing $f^{-1}$ is not continuous.  However, I am confident in my second example.
99741
Given that $X$ is an arbitrary topological space, $Y$ is a totally ordered set in the order topology, and $f$, $g \colon X \to Y$ are continuous functions, how to show that the subset $A$ of $X$ given by  $$ A \colon= \left\{ \ x \in X \  \colon \  f(x) \leq g(x) \ \right\} $$  is closed in $X$?   Edit based on the answer by Hagen von Eitzen:   Let $U \colon= \{ u \times v \in Y \times Y \ \colon \ u &lt; v \ \}$. We show that $U$ is open in $Y \times Y$.   Let $u \times v \in U$. Then $u &lt; v$.   Case I. If there is some $y \in Y$ such that $u &lt; y &lt; v$, then $u \times v \in (-\infty, y) \times (y, +\infty) \subset U$.   Case 2. If $(u,v)$ is empty, then $u \in (-\infty, v)$ and $v \in (u, +\infty)$, and so $u \times v \in (-\infty, v) \times (u, +\infty)$.   Moreover, if $a \times b \in (-\infty, v) \times (u, +\infty)$, then we must have $a &lt; v$ and $b&gt; u$. So $a \leq u &lt; v \leq b$, which implies that $a &lt; b$ and so $a \times b \in U$.   Thus, $u \times v \in (-\infty, v) \times (u, +\infty) \subset U$.   So $U$ is open in $Y \times Y$.   Similarly, we can show that the set  $$V \colon= \left\{ \ u \times v \in Y \times Y \ \colon \ u &gt; v \ \right\}$$ is open in $Y \times Y$.   Thus, it follows that the set $A \colon= \{ \ u \times v \in Y \times Y \ \colon \ u \leq v \ \}$ is closed in $Y \times Y$.   Now since the maps $f \colon X \to Y$ and $g \colon X \to Y$ are continuous, so is the map $f \times g \colon X \to Y \times Y$ defined as $$(f \times g)(x) \colon= f(x) \times g(x) \ \mbox{ for all } \ x \in X.$$  Thus, the inverse image under $f \times g$ of the set $A$ is closed in $X$.   But  $$  \begin{align}  (f \times g)^{-1} (A) &amp;= \left\{ \ x \in X \ \colon \ (f \times g)(x) \in A \ \right\} \\ &amp;= \left\{ \ x \in X \ \colon \ f(x) \times g(x) \in A \ \right\} \\ &amp;=  \left\{ \ x \in X \ \colon \ f(x) \leq g(x) \ \right\}. \end{align}  $$
79243
Suppose $f:X \to Y$ is a continuous proper map between locally compact Hausdorff spaces. Are the following results true?  $1$. The map $f$ takes discrete sets to discrete sets.  $2$. If $f$ is injective, then $f$ must be a homeomorphism onto its image.  Edit1:Let $A$ be discrete in $X$ and let $K$ be compact in $Y$ then $f(A) \cap K=f(A \cap f^{-1}(K))$,is finite since $A \cap f^{-1}(K)$ is finite.Hence $f(A)$ is discrete. As there is a counterexample in answer below so can someone please point out the error in my proof?
53383
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.
81589
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?
88078
Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, is $H(C)$ necessarily dense in the $Y$?  I thank you in advance.
96000
Let $I$ be the unit interval, and define the torus using the usual identifications on $I \times I$. I've shown that $I \times I - \{x\}$ (where $x$ is a point not on the boundary of $I \times I$) is homotopy equivalent to its boundary (and that it even retracts onto its boundary). Let $p: I \times I \rightarrow I \times I /\sim$ be the canonical quotient map. Using the retraction above, call it $r$, is there a way I can explicitly show that $p(I \times I - \{x\})$ is homotopy equivalent to $p(\text{boundary of } I \times I)$, i.e., that torus minus a point to a figure-eight?  More generally, if $A \subset X$, and $X$ is homotopy equivalent to $A$ (or $X$ retracts to $A$), for which maps $p$ can I say that $p[X]$ is homotopy equivalent to $p[A]$?   Hints appreciated!
8901
, it is said that a space-filling curve cannot be injective because "that will make the curve a homeomorphism from the unit interval onto the unit square", since every continuous bijection from a compact space to a Hausdorff space is a homeomorphism.  That does mean there can be no injective mapping from $[0,1]$ onto $[0,1]\!\times\![0,1]$, but it doesn't really tell us anything about mappings from noncompact intervals onto the unit square. In the case of the ordinary Peano-Hilbert curve, it does not help much: the construction has infinitely (only countably, though) many inherent points of overlap (discussed here:  ), but I can't really see any reason why there could not be a space-filling curve without such points. Of course, the curve's inverse could not be continuous, but when mapping from a noncompact interval it wouldn't need to be! After all, there are other curves mapping noncompact spaces bijectively to compact ones in a continuous way, but with noncontinuous inverse, like the obvious $[0,2\pi[\to S^1$.
93443
$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$  I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.
44708
Suppose we have a bijective continuous map $\mathbb{R}^n\to\mathbb{R}^n$ (relative to the standard topology). Must this map be a homeomorphism?  I have little doubt about this. I think that if it happens, I guess it's true, I've heard it is true, but I can not prove it.
63788
Let $X$ be a non-empty set. $N(X)$ denotes the collection of all nets on $X$ and $\mathcal{P}(X)$ denotes the collection of all subsets of $X$. Let $T:N(X)\to \mathcal{P}(X)$ be a map satisfying the condition $T\{x_\lambda\}_{\Lambda}=T\{y_\gamma\}_{\Gamma}$ if one of $\{x_\lambda\}_{\Lambda}$ and $\{y_\gamma\}_{\Gamma}$ is a subnet of the other. Define $\tau=\{A\subset X|\ T\{x_\lambda\}_{\Lambda}\subset A^c\text{ for every net }\{x_\lambda\}_{\Lambda}\subset A^c\}$  We have to prove that $\tau$ is a topology on $X$. I have tried the problem in the following manner-  $\emptyset\in \tau$ since given any net $\{x_\lambda\}_{\Lambda}\subset (\emptyset)^c=X$, we always have $T\{x_\lambda\}_{\Lambda}\subset (\emptyset)^c=X$ . Again, $X\in \tau$ since $X^c=\emptyset$ Let $\{A_i|\ i\in I\}\subset \tau$. Then let $\{x_\lambda\}_{\Lambda}\subset \left(\bigcup A_i\right)^c=\bigcap A_i^c\implies \{x_\lambda\}_{\Lambda}\subset A_i^c\ \forall i\implies T\{x_\lambda\}_{\Lambda}\subset A_i^c\ \forall i\implies T\{x_\lambda\}_{\Lambda}\subset \bigcap A_i^c=\left(\bigcup A_i\right)^c$. Hence, $\bigcup A_i\in \tau$. Let $A,B\in \tau$ and $\{x_\lambda\}_{\Lambda}\in (A\cap B)^c=A^c\cup B^c$. Now we have to prove $T\{x_\lambda\}_{\Lambda}\subset A^c\cup B^c$. Now I define $\Gamma=\{\lambda\in\Lambda|\ x_\lambda\in A^c\}, \Sigma=\{\lambda\in\Lambda|\ x_\lambda\in B^c\}$. Now if I can prove at least one of $\Lambda_1,\Lambda_2$ is directed set then I will construct a subnet of $\{x_\lambda\}_{\Lambda}$, and use the property of $T$ to achieve our goal. But I'm to prove that at least one of $\Gamma,\Sigma$ is directed set, although I have shown that both $\Gamma, \Sigma$ satisfies reflexivity and transitivity condition. But we have to show the third condition which is $$\text{If }\lambda_1,\lambda_2\in \Gamma\text{ (or }\Sigma\text{)}, \exists \lambda_3\in \Gamma\text{ (or }\Sigma\text{) such that } \lambda_3\ge\lambda_1,\lambda_2$$  Can anyone help me with this regard? Thanks for your help in advance.
92801
I want to ask about the homeomorphism between $\mathbb{Q}$, $\mathbb{Q}_{&gt;0}$: the rationals greater than $0$, and $\mathbb{Q}_{\geqslant 0}$: the rationals $\geqslant 0$?  For $\mathbb{Q}$ and $\mathbb{Q}_{&gt;0}$, I can use the back and forth map to have an order-isomorphism, and as a result, a homeomorphism. Is there a direct formula for the map?  It's very surprising that $\mathbb{Q}_{\geqslant 0}$ is homeomorphic to $\mathbb{Q}_{&gt;0}$ and $\mathbb{Q}$,  I don't know how to show they are homeomorphic.    Could anyone please help me with the homeomorphisms between them?
75149
I hope this doesn't turn out to be a silly question.  There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.    My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?    For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.
85871
Working on this problem.      If any $f:S^1\rightarrow X$ extends to a $F:D^2\rightarrow X$, then $\pi_1(X, x_0)$ is trivial.    We can turn any loop $f^\prime:I \rightarrow X$ in $\pi_1(X, x_0)$ into an $f:S^1\rightarrow X$ by defining $f(x) =f^\prime \left(\frac{1}{2\pi}\operatorname{Arg}(x) \right) $.   I want to finish without explicitly showing that $f$ is homotopic to a constant map. (This is part of the "the following are equivalent proof" with that as part (a) so I would like to avoid assuming that or proving it midway. I should be able to do it with the extension map only.) I think I am supposed to use induced maps here and make some kind of diagram but I'm not sure how.
55332
Let $X$ be a topological space and let $\Delta = \{(x,x) : x\in X \}$ be the diagonal of $X\times X$ (with the product topology).     I was asked to prove that $X$ is ${\rm T}_1$ if and only if $\Delta$ can be written as intersection of open subsets of $X\times X$.   I think is better (maybe easier) to use the well-known result "$X$ is ${\rm T}_1$ if and only if $\{x\}$ is a closed set $\forall x\in X$".  What I've done:  Assuming that $X$ is ${\rm T}_1$, let $Y=(X\times X) \setminus\Delta$ and note that (trivially) $$Y = \bigcup_{y\in Y} \{y\}.$$ Then $$\Delta = (X\times X) \setminus Y = (X\times X) \setminus \left(\bigcup_{y\in Y} \{y\} \right) = \bigcap_{y\in Y} (X\times X)\setminus \{y\},$$ where $(X\times X)\setminus\{y\}$ is open since each $\{y\}$ is closed.  The other direction seems be a bit harder, I've tried unsuccessfully. Any ideas?  Thanks in advance.
79522
I was wondering whether $M\times \mathbb{R}$ is homeomorphic to $N\times \mathbb{R}$ implies $M$ is homeomorphic to $N$, where let us say $M,N$ are smooth manifolds. (They are certainly homotopy equivalent.) More generally what if $\mathbb{R}$ is replaced by some other manifold say $\Sigma$? Any possible approach/idea?