I know $D\subseteq Y$ is dense $\implies \overline{D}=Y$ or $D\cap U\neq\emptyset$ $\forall$ $U$ open. Also if $f$ is continous then $f^{-1}[U]$ is open. I suppose that $f^{-1}[D]\cap f^{-1}[U]\neq\emptyset$ may be false. I don't know how an open map takes part on the proof.
15979
Let $X,Y$ be topological and $f:X \to Y$ be open and continuos.  I am studying Baire space and I would like to try the following facts :      $(i)$ The inverse image of dense set is dense.       $(ii)$ The inverse image of comeager set is comeager.   I am studying Baire space and I would like to try the following facts  Can we collaborate to verify the above claims?    Thanks.
42031
I'm looking for proof verification/help for the title question. Here is what I have now: Let $f:X \rightarrow Y$ be continuous with $X$ limit point compact. Let $V$ be an infinite subset of $f(X)$. Then we also have that $f^{-1}(V)$ is also an infinite subset of $X$. Since $X$ is limit point compact there is $x \in X$ such that $x$ is a limit point of $f^{-1}(V)$. Then $x \in \overline{f^{-1}(V)}$ by definition of the closure. Hence since $f$ is continuous, $$ f(x) \in f(\overline{f^{-1}(V)}) \subseteq \overline{f(f^{-1}(V))} \subseteq \overline{V} $$ Then since $f(x) \in \overline{V}$, it is either a limit point of $V$ or an element of $V$. However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case. Thus $V$ has a limit point and $f(X)$ is limit point compact.
4569
Question:  Let $f: \mathbb R \to \mathbb R$ be continuous and open, that is if $A \subset \mathbb R$ is open then $f(A) \subset \mathbb R$ is open. Prove that $f$ is injective.   Attempt:  Suppose $f$ is not injective then there exit $x, y \in \mathbb R$ such that $$x &lt; y \implies f(x) = f(y) = c$$  Take the closed inteval $[x,y] \subset \mathbb R$, as $f$ is continuous then $\displaystyle {f|_{[x,y]}}$ is continuous, by Weierstrass Theorem we have that $f$ has a maximum or a minimum point.   Let's assume $m = \max \{f(a) ; a \in [x,y]\}$. Now there is  $x' \in [x,y]$ such that $f(x') = m$. If we take the open $(x'-\delta, x'+\delta)$ centered at $x'$ and  we have  (1)  If $m = c$ then $f$ is constant on the interval $[x,y]$ and $f((x'-\delta, x'+\delta)) = \{c\}$ which is closed, thus a contradiction;   (2)  If $m \neq c$ then we would have $f((x'-\delta,  x'+\delta)) = (b, m]$, where $b$ can also be $b = \infty$. Again a contradiction.   Well, this is my least embarassing attempt. I'm not sure how to show $(2)$ $100 \%$.   I have also tried to show $f^{-1}f(A) = A$ for any $A \subset \mathbb R$, tried to work on the connected space  $\mathbb R$ by finding a contradiction using the intervals $E_{[f &gt; c]}$ and  $E_{[f &lt; c]}$ open when $A$ is open.   Any thoughts?  Note: I've already seen  to try something out, but the fact that $f$ is monotone on this exercise comes as a consequence.
60367
I was wondering if this proof was right.  $\Leftarrow$ Suppose $E$ is closed. Then choose $x\in E^{c}$, then $x\notin E$, and so $x$ is not a limit point of $E$.   Hence there exists a neighborhood $N$ of $x$ such that $E \cap N$ is empty, such that $N \subset E^{c}$. Thus $x$ is an interior point of $E^{c}$ and $E^{c}$ is open  $\Rightarrow$ Then suppose $E^{c}$ is open. Let $x$ be a limit point of $E$, then every neighborhood of $x$ contains a point of $E$, so that $x$ is not an interior point of $E^{c}$. Since $E^{c}$ is open, this means that $ x \in E$, Therefore $E$ is closed.
101938
Let $X$ be a topological space and suppose that $H$ and $G$ are open dense subsets of $X$.Then show that $G \bigcap H$ is also an open dense subset of $X$.  My attempt :  Well since the finite intersection of open sets is open therefore, $G \bigcap H$ is also open.  I was trying to prove the other part by the method of contradiction :  Suppose on the contrary that $G \bigcap H$ is not dense. Then that implies that $Int(Cl(G \bigcap H)) = \phi$ .  Now, $Cl(G \bigcap H) \subset Cl(G) \bigcap Cl(H)$  so, $Int(Cl(G \bigcap H)) \subset Int(Cl(G) \bigcap Cl(H)) = Int(Cl(G)) \bigcap Int(Cl(H)) = X $.  How do i proceed further to get a contradiction ?
99142
Let $X$ and $Y$ be two topological spaces.   Prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.   I am stuck on the converse. Suppose $f: X \to Y$ is continuous. Then for every closed set C in Y, $f^{-1}(C)$ is closed in X.   WTS  for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.     Since $\overline{f(A)}$ is a closed set in Y, $f^{-1} \circ \overline{f(A)}$ = $ \overline{ f^{-1} \circ \overline{f(A)} }$  Also, $f(\overline{A}) \subseteq \overline{ f(\overline{A})}$
79839
Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)&lt;\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.  The second assertion is easier since the quotient space $X/N$ has the universal mapping property.
36055
This problem was posted before, but not the proof (because the asker knowed the answer), only a counterexample without the hypothesis of finite fibres.  I want to know how to prove this proposition:  Let $q:X\to Y$ and $r:Y\to Z$ be covering maps. Let's suppose that for each $z\in Z$ , the set $ r^{-1}(z)$ is finite, then the composition $p = r\circ q$ is also a covering map. Well I have to consider $z\in Z$ and show that there exist a neighborhood that is evenly covered by $p$, I know that there exist a neighborhood $U$ that is evenly covered by $r$, I think that this will be the desired neighborhood. First of all $ r^{-1}(U) = \cup_{i=1}^{n}  V_i $ (it's easy to see that the unions is finite using the fact that the fibres are finite and this is a local homeomorphism between the $V_i$) Then $ p^{-1}(U)=q^{-1} (r^{-1}(U) ) = q^{-1}(\cup_{i=1}^{n}  V_i)=\cup_{i=1}^{n}  q^{-1}(V_i) $ With $ q^{-1}(V_i) \cong V_i$ under $q$ and so $\cong$ U under $p$ I never used the fact that the preimage is finite, so my proof is obviously not correct, please help me with this
76671
I'm in search of different proofs of the following proposition:     $\bf{Proposition}$: Suppose $X$ and $Y$ be topological vector spaces, $\text{dim }Y&lt;\infty$, and $\Lambda:X\to Y$ is a surjective linear map. Then $\Lambda$ is open.   Any and all proofs are welcomed.
47578
Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, \end{align*} for all $x \in [0,1]$. Why is $E_n$ nowhere dense in $C\big([0,1]\big)$.  I have been able to show that $E_n$ is closed in $C\big([0,1]\big)$, but haven't been able to show why the interior is empty. Any advice?...
77246
Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a continuously differentiable map satisfying $\|f(x)-f(y)\|\geq\|x-y\|$ for all $x,y\in\mathbb{R}^{n}.$ Then   $A.$ $f$ is onto.  $B.$ $f(\mathbb{R}^{n})$ is a closed subset of $\mathbb{R}^{n}.$  $C.$ $f(\mathbb{R}^{n})$ is a open subset of $\mathbb{R}^{n}.$  $D.$ $f(0)=0.$  According to me $f$ is one one and onto so image is closed and open both. Please suggest me. Thanks in advance.
16363
Hello I am doing this exercise where we let X and Y be topological spaces and $f:X\to Y$ a function. We suppose that $\mathcal{A}$ is a collection of closed subsets of X, such that $\bigcup_{A\in\mathcal{A}} A = X$. We assume that for every $A\in \mathcal{A}$ the restriction $f_{|A}: A\to Y$ of f to A is continuous. Furthermore we suppose that for each $x\in X$ there exists a neighborhood U of x such that the intersection $A\cap U$ is non-empty for only finitely many $A\in\mathcal{A}$. And I have to show that f is continuous. I know that since $f_{|A}: A\to Y$ is continuous we have that for all open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of A. And to show that $f: X\to Y$ is continuous I am thinking of showing that for each open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of X, but this would be obvious wouldn't it since for all open $O\subseteq Y$ we have $f^{-1}(O)\subseteq A\subseteq X$?
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 $Y$?
74019
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?
24144
I was assigned the following homework problem for an introductory course in topology:  Let $p:\ X\to Y$ be a closed continuous surjective map such that $p^{-1}(\{y\})$ is compact for each $y\in Y$.  Show that if $Y$ is compact, then $X$ is compact.  My question is, are the assumptions that $p$ be continuous and surjective even necessary for the result to hold?  I've done the following:  Let $U$ be an open set containing $p^{-1}(\{y\})$.  Since $X-U$ is closed, $p(X-U)$ is closed in $Y$.  Then $W=Y-p(X-U)$ is an open set that contains $y$.  Since $(X-U)\cap p^{-1}(W)=\varnothing$, $p^{-1}(\{y\})\subset p^{-1}(W) \subset U$.  Let $\{U_\alpha\}$ be an open cover of $X$.  Then $p^{-1}(\{y\})\subset \bigcup_{k=1}^{N}U_k$.  For each $y\in Y$, define $W_{y}$ as above; then $\{W_y\}$ is an open cover of $Y$, so can be covered by $W_{y_1},...,W_{y_m}$, and $p^{-1}(\{y_j\})\subset p^{-1}(W_{y_j})\subset \bigcup_{k=1}^{N}U_k$.  Since $X=p^{-1}(Y)$, $X=\bigcup_{j=1}^{n} p^{-1}(W_{y_j})$.  Since each $p^{-1}(W_{y_j})$ is covered by finitely many $U_k$, so is $X$, so it is compact.  If we dispense with surjectivity of $p$, the only thing that happens is that $p^{-1}(\{y\})$ may be empty, but then $p^{-1}(\{y\})\subset p^{-1}(W)$ holds vacuously, and the empty set is still compact.  If $p$ is continuous, then $p^{-1}(W)$ is open, but I don't use the fact that it is open at all in my proof.  I know that all 3 assumptions make a map "perfect" so that may be why they were mentioned, but can we dispense with continuity and surjectivity?  Thanks.
5773
The  states that, given an open subset $U\subseteq \mathbb{R}^n$ and an injective and continuous function $f:U\rightarrow\mathbb{R}^n$ then $f$ is a homeomorphism between $U$ and $f$'s image.   I tried proving it by using another theorem:  if $g:K\rightarrow X$ is injective and continuous, $K$ is compact and $X$ is  then $g$ is a homeomorphism between $K$ and $f(K)$.  But I'm not sure on how to prove this (sub)-theorem? or perhaps there exists an easier proof of the invariance of domain theorem?
20353
There was an exercise I could not do. So the property is $T_1$ if for every pair of distinct points, $P, Q \in X$, there is an open subset $U$ containing $P$ but not $Q$ and another open subset $V$ containing $Q$ but not $P$.  So I was asked to show if the base field is infinite, then $\mathbb{A}^n$ is $T_1$ but not Hausdorff.  I can show that it's not Hausdorff (I believe this is right). So let $k$ be an infinite field, then we have $U$ and $V$ which are two nonempty subsets which are open, thus $U^c$ and $V^c$ finite so by Demorgan's law we have $(U \cap V)^c =U^c \cup V^c$ which is also finite, so this implies that $U\cap V$ is non-empty so it can't be Hausdorff.  But I can't seem to really show that it's $T_1$  Any ideas how to approach this?  Thank you.