Lemma:   Let $f:(X,\mathcal{T})\longrightarrow(X',\mathcal{T'})$ be a mapping  between topological spaces where $$X=\cup_{j=1}^m F_j$$   $ \text{ with }F_j\text{ closed for }\mathcal{T},\, f_{|F_j} \text{ continuous },\forall j=1,\ldots,m.$   Then $f$ is continuous.     I'd like to find a non-example to that when $X$ is a countably infinite union of closed sets. My course already provides one that I find counterintuitive as I've mostly worked so far on metric spaces (the identity between $\mathbb{N}=\cup\{n\}$ with the cofinite topology and $\mathbb{N}$ with the discrete topology). Is it possible to find one where both spaces are $\mathbb{R}$ with the standard topology ? I've considered disjoint closed $X_n$ that increasingly cover $\mathbb{R}$ but in vain.
2004
The gluing lemma for closed sets states:     Given a finite closed cover $\{A_i\}$ of a topological space $X$, together with continuous maps $\{f_i : A_i \to Y\}$ into some other topological space $Y$, there exists a unique continuous map $f : X \to Y$ whose restriction to each $A_i$ is equal to $f_i$.   Question: What is a good and simple counter-example when the gluing lemma fails in the case that $\{A_i\}$ is infinite, but countably so.  My attempt: I have only been able find a silly counter-example: let $\{A_\alpha\}$ be the collection of all points of $X = [0,1] \subset \mathbb{R}$, and let $f_\alpha = 0$ for all $\alpha$ except for $\alpha = \alpha_0$, for which $A_{\alpha_0} = \{0\}$ and $f_{\alpha_0} = 1$.
99801
How to show that $(1)\Longleftrightarrow (2)$ in metric spaces ?   pre-image of open sets are open   $\delta$-$\epsilon$ definition of continuity
72882
Problem:      Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y  $ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)  $ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)  = g(x) $ for all $x \in  X $.   Attempt at a solution:  Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) &gt; 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y $ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.   I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A $, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.  Also, Would be be possible to prove this without using the Hausdorff condition on $Y$?
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.
62438
Here is the question:     Prove that the function $f: E \rightarrow E'$ between metric spaces is continuous iff whenever $ C\subset E'$ is closed, $f^{-1}(C)$ is a closed subset of $E$.   What I was thinking of doing is this following:  First, assume F is continuous. So we want to show that whenever  $ C\subset E'$ is closed, then $f^&lt;(C)$ is a closed subset of $E$. To do this, do we just need to show that whenever $C^c \subset E'$ is open, then $f^&lt;(C^c)$ is open?
41208
This problem is bugging me for some time.     $f:(X,\mathcal{T}) \to (Y,\mathcal{T}')$ is a continuous bijection where $(X,\mathcal{T})$ is compact and $(Y,\mathcal{T}')$ is $T_2$ (i.e. Hausdorff and $T_1$). Then show that $f$ must be a homeomorphism.   So far I have failed to understand any relation between the compactness of domain and $T_2$-ness of the range. I tried in vain to prove that $f(U)$ is open in $(Y,\mathcal{T}')$ whenever $U$ is so in $(X,\mathcal{T})$. I know $f$ makes $X$ and $Y$ respectively $T_2$ and compact and hence makes both of them $T_4$ spaces. But that is not coming to any use. I would be glad if someone helps me with a solution. Thank you.
63782
Let $X$ be a Hausdorff topological space , let $f:X \to X$ be a continuous map such that $f\circ f=f$ , then is it true that $f(X)$ is closed in $X$ ?
121049
Show that for any topological space $X$ the following are equivalent:         $X$ is extremally disconnected   Every two disjoint open sets in $X$ have disjoint closures.      My attempt at a solution  Assume every two disjoint open sets in $X$ have disjoint closures. Let A open in $X$, $A$ and $X \setminus \overline{A} $  disjoint open set $\overline{A} \cap (\overline{X \setminus \overline{A}} )= \emptyset$,  $(\overline{X \setminus \overline{A}} ) \subset X \setminus \overline{A}$, which implies that $X \setminus \overline{A}$ is closed in $X$. Thus $\overline{A}$ is open in $X$. So we obtain $X$ is extremally disconnected.  Otherwise Let $X$ is extremally disconnected. Then the closure of every open set is open.  So $X$ is completely separated. Let $A$, $B$ two disjoint open sets.  Then there exist a continuous function $g:X \rightarrow [0,1]$ such that  $g(A)=0$, $g(B)=1$  How can I continue?
12920
Let $X$ be a metric space.  If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.  I'm looking for a direct proof of this theorem.  (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.")  Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.  Here are the definitions I'm using:  Let $X$ be a metric space with distance function $d(p, q)$.  For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) &lt; r\}$.  Any $p \in X$ is a limit point of $E$ if $\forall r &gt; 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$.  Any subset $E$ of $X$ is closed if it contains all of its limit points.
117942
For a closed subset $Y$ of a space $X$ we have the following inequlity of topological (covering) dimensions: $$\dim{Y} \leq \dim{X}$$ (assuming at least one of those is finite).  I have two questions related to that:   Does the above inequality hold for any (not necessarily closed) $Y \subseteq X = \mathbb{R}^n$? Can you come up with some nice counterexamples to the above inequality when we skip the assumption that $Y$ is closed? By nice I mean something nicer than my counterexample described below.   Not nice counterexample  I'm not happy with it, because it is not even a Hausdorff space and it is quite artificial for me.  Let $C_n=[n-1,n] \times [0,1]^{n-1}$ ($C_1=[0,1]$). Then $D_m = \bigcup_{n=1}^m C_n$ (for $m \in \mathbb{N} \cup {\infty} $). Define the topology $\tau_m$ on $D_m$ with a normal euqlidean metric on $D_m \subseteq \mathbb{R}^m$. For $m&lt;\infty$ $D_m$'s dimension is $m$ since it is a finite sum of closed subsets $C_n$ and maximum of their dimensions is $m$*. For $m=\infty$ we can notice that $D_N \subseteq D_m$, $D_N$ is closed and its dimension is $N$ and then go with $N$ to infinity.  The exact counterexample:  (1) We can define $(D&#39;_m, \tau&#39;_m)$ as $(D_m \cup \{Inf\}, \tau_m \cup \{D&#39;_m\})$. Then its dimension is zero since the only open set containing $Inf$ is the whole space so it must be included in any open covering. However zero-dimensional $D&#39;_m$ contains $D_{m&#39;}$ for any $m&#39; \leq m$, which dimension is $m&#39; \gt 0$.  (2) Moreover one can define more additional (containing $Inf$) sets in topology in such a way that they don't have to contain the whole $D_m$ but only its "tail" without some initial part $D_{m&#39;} \subsetneq D_m \subseteq D&#39;_m$ ($m' \lt m$) and then the dimension of the whole space would be $m&#39;$.  *I haven't seen the proof that $\dim{\mathbb{R}^n}=n$, but I strongly believe in it.  EDIT:  As Brian points below, my inequality is not true with Engelking's definition of $\dim$ with functionally open sets. The definition that I meant is the following (X is any topological space):     A space X is said to be finite dimensional if there is some integer $m$ such that for every open covering $A$ of X, there is an open covering $B$ of X that refines $A$ and has order at most $m+1$. The topological dimension of X is defined to be the smallest value of $m$ for which this statement holds; we denote it by $\dim$X.   (Munkres J.R., Topology, paragraph 50)  I'm sorry for the lack of precision.
89878
I am struggling with an exercise in Topology. The exercise goes:  Let $Y$ be Hausdorff, and $X\subseteq Y$ an open subspace which is locally compact. Let $X^{+}$ denote the one-point compactification of $X$, and let $\infty$ denote the unique point in $X^{+} \backslash X .$ Define $f: Y \longrightarrow X^{+}$ by $$ f(y)=\left\{\begin{array}{ll} y &amp; \text { if } y \in X \\ \infty &amp; \text { else } \end{array}\right. $$ Show that $f$ is continuous.  I have a lot of definitions and lemmas about continuity, but none of them seems to fit with this problem. I am looking for a hint on how to get started with the proof. Thanks in advance.
16420
My attempt to prove the following is below:     Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$   My book calls this set the ""  I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly)  Using the definition of topology:     $\mathscr{T}$ is a topology on X if and only if the following are true:   (i) X and Ø are elements of $\mathscr{T}$.   (ii) $\mathscr{T}$ is closed under finite intersections.   (iii) $\mathscr{T}$ is closed under arbitrary unions.   Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2...$ are all finite sets)  So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.   Now to verify the definition.     (i) X and Ø are elements of $\mathscr{T}$.   Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.      (ii) $\mathscr{T}$ is closed under finite intersections.   $U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| &lt; |\mathbb{N}|  $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.       (iii) $\mathscr{T}$ is closed under arbitrary unions.   $U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| &lt; |\mathbb{N}|  $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.  Now let's consider the infinite case: $V_k \cap ... \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap ... $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.
41619
I'm self studying J. Lee's Introduction to Topological Manifolds, and after doing all other exercises on chapter 2, I can't seem to come with the proper counterexamples for this one.      For each of the following properties, give an example consisting of two subsets $X, Y \subset \mathbb{R}^2 $, both considered as topological spaces with their Euclidean topologies, together with a map $f: X \rightarrow Y$ that has the indicated property.         Open but not closed/continuous   Closed but not open/continuous    Continuous but not open/closed         etc... (all cases)   I think the reason for my difficulty is a lack of geometric intuition of what open, closed and continuous maps look like as deformations of shapes in $\mathbb{R}^2$ (I can construct counterexamples from artificial made up spaces).   I tried squashing balls and 'pumping space' into lines, as well as joining strips in annulus shapes, but all the examples I seem to come up with are either too well behaved (i.e. all 3) or too misbehaved.   Instead of a case by case answered I would prefer a couple worked out cases and some intuition on how to visualise open / closed maps in this context.
57407
It is a Dugundji's exercise:  Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.  If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.  For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$.   Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.
50097
I have $A$ as an infinite set and $S$ as a countably infinite set, (so that means there exists a one-to-one correspondence between $S$ and $\mathbb{N}$).  How do I show that there always exists a bijection between $A$ and $A\cup S$? Can it be done by showing that there is a bijection from $A$ to $S$ or from $A$ to itself? I am lost on this one.   Oh, and can it be possible that there is no bijection if it is between $S$ and $A\cup S$? What about a map that maps $\mathbb{N}\to \mathbb{N}\cup\mathbb{R}$?
83281
Following a reference from "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.     Since the sum and the product of real continous functions, defined in any topological space, are continuous we can easily prove that every metric space is a Tychonoff spaces. So we suppose that $(X,d)$ is a metric space and we consider a closed set $F$ a point $p\notin F$ and so we define a function $f:X\rightarrow[0,1]$ by the condiction    $$ f(x)=\frac{d(x,p)}{d(x,p)+d(x,F)} $$   where $d(x,F):=\mathscr{inf}\{d(x,a):a\in F\}$. So we observe that $f(p)=0$ and $f(x)=1$ for any $x\in F$; then $f$ is continuous, since any function $g:X\rightarrow\mathbb{R}$ defined as    $$ g(x)=d(x,A) $$       for any $x\in X$ and for any fixed $A\subseteq X$, is a countinuous function and so $f$ is continuous  function, since the sum and the ratio of continuous function is a continuous function.       Well by definition of Tychonoff space we can claim that the metric space $(X,d)$ is a Tychonoff space.   Well unfortunately I don't be able to prove that the function $g$ is continuous: could someone help me, please?
6038
Let $X$ and $Y$ be metric spaces, $X$ compact, and $T:X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism.   My attempt:  We need only show that $T^{-1}$ is continuous.  Let $M \subset Y$ be closed. Then, from the surjectiveness of $T$, there exists a $A \subset X$ such that $T(A)=M$. Since $M$ is closed, by the continuity of $T$, we must have that $A$ is closed.  Now $$T^{-1}(T(A))=T^{-1}(M)$$ i.e, $A=T^{-1}(M)$. Since $A$ and $M$ are closed, we must have that $T^{-1}$ is continuous.  My problem with this, however, is that I am not using the fact that $X$ is compact anywhere in the proof. Is there perhaps another way that I can show this, using the compactness of $X$?   Any hints will be appreciated.