Let $p:X \rightarrow Y$ be a perfect map, that is, $p$ is closed, continuous, surjective, and $p^{-1}(y)$ is compact for every $y \in Y$. Show that if $X$ is locally compact, then so is $Y$.  There are a couple of these perfect map problems on one of our old qualifying exams. They seem really difficult. I've been working on this one all night and can't figure it out.
122542
Recall that a space $X$ is locally compact if for any point $x$ in $X$, and any neighborhood $U$ of $x$ in $X$,there is a neighborhood $V$ of $x$ in $X$,and a compact subspace $C$ of $X$ such that $x \in V \subset C \subset U.$ Let $X$ be a locally compact space, and let $f : X \rightarrow Y$ be a continuous, closed surjection. Suppose that $f^{-1}(y)$ is compact for each $y \in Y$ . Show that $Y$ is locally compact.
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.
44856
I am trying to prove the following:  "Let $U$ be an open subset in the compact metric space $X$. If $\{S_k : k\in\mathbb{N}\}$ is a collection of closed subsets of $X$ such that $S_{k+1}\subset S_k$ for all $k\in\mathbb{N}$, and $\bigcap_{k=1}^{\infty} S_k \subset U$, then $S_n\subset U$ for some $n\in\mathbb{N}$."  Here is what I have so far:   Being closed subsets of a compact set, each $S_k$ is compact.   Additionally, I have a theorem which states that an infinite collection of nested compact subsets of $X$ has a nonempty intersection. Therefore, $\bigcap_{k=1}^{\infty} S_k \subset U$ implies that there exists an element $p$ such that $p\in\bigcap_{k=1}^{\infty} S_k \subset U$, i.e., each $S_n$ contains at least one point which is an element of $U$. Being open, there exists some $\epsilon &gt;0$ such that $B(p,\epsilon)\subset U$ (I'm not sure if that is important).  I haven't been able to get any further. This problem is given not long after first being introduced to compact sets, so I don't expect the completion of the proof will require much theory. Any hints or a solution or a point in the right direction would be greatly appreciated. Thank you.   Also, I am sorry if this is a duplicate, I haven't been able to find anything like it. I am also not sure if the (analysis) tag is appropriate, I don't know of a more specific tag which is relevant.
117041
Let (X,d) be a compact metric space. For every open cover, show there exists ε > 0 such that for every x ∈ X, B(x,ε) is contained in some member of the cover.   My attempt:  (X,d) is compact. Therefore there exists a finite subcover of X.  Any element x in X must lie in some member of the cover, say x ∈ Ui. Otherwise they would not constitute a cover.  Since Ui is open, by definition every point is interior, so there exists ε > 0 such that B(x,ε) is contained in Ui.   I haven't used the fact the subcover is finite, or the fact X is a metric space rather than just topological space, so I feel my reasoning is flawed.   Any help is greatly appreciated!
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?
3875
I'm really confused with the statement.   let $X$ be a set endowed with the discrete topology and let $Y$ be any topological space. Show that $M(X,Y)$, the space of continuous maps from $X$ to $Y$ endowed with the compact open topology, is homeomorphic to the topological product $ \prod_{x \in X} Y_x$, $Y_x = Y$.  Can you please explain what we really have to prove?
74221
Let $\emptyset \not = X\subseteq \Bbb{R}^n$ be convex and compact and let $\cal{A}$ be a commuting family of affine maps from $\Bbb{R}^n$ into $\Bbb{R}^n$ such that $X$ is invariant under each element of $\cal A$. How can I show that $\cal{A}$ has a common fixed point.   Remark I know this is a consequence of some stronger theorems such as Markov-Kakutani fixed point theorem. But I have no access to its proof. Besides I am looking for a much elementary proof (which I think must exists for this special case).
120483
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.
2650
Theorem. Let $X$ and $Y$ be sets with $X$ nonempty. Then (P) there exists an injection $f:X\rightarrow Y$ if and only if (Q) there exists a surjection $g:Y\rightarrow X$.  For the P $\implies$ Q part, I know you can get a surjection $Y\to X$ by mapping $y$ to $x$ if $y=f(x)$ for some $x\in X$ and mapping $y$ to some arbitrary $\alpha\in X$ if $y\in Y\setminus f(X)$. But I don't know about the Q $\implies$ P part.  Could someone give an elementary proof of the theorem?
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 $f^{-1}(O)\subseteq A\subseteq X$?
86563
It seems as if every subspace of a compact topological space (equipped with its relative topology) had ought to be compact as well. Is this true in general?  And in particular, I want to use the fact that every quasi-projective variety is compact with the Zariski topology - I remember having proven it for quasi-affine varieties, but the same proof does not work for quasi-projective varieties. Is every quasi-projective variety compact?
105127
if $X$ and $Y$ are Hausdorff spaces, $f:X \to Y$ is a local homeomorphism, $X$ is compact, and $Y$ is connected, is $f$ a covering map?  It seems to be, and I almost have a proof, but I'm stuck at the very end of it:  I've already proved that $f$ is surjective (using the connectedness), and that for each $y \in Y$, $f^{-1}(y)$ is finite. Because $X$ is compact, there exists a finite open cover of $X$ by $ \{ U_i \}$ such that $f(U_i)$ is open and $f |_{U_i}:U_i \to f(U_i) $ is a homeomorphism. For each $y \in Y$, we choose the subset $ \lbrace U_{i_j} \rbrace $ such that $y \in U_{i_j}$, and then define $V = \bigcap_{j=1}^k f(U_{i_j})$, and $U'_j = U_{i_j} \bigcap f^{-1}(V)$.  ... and this is were I got stuck. I really want to write that $f^{-1}(V) = \bigcup_{j=1}^k U'_j$ (more or less proving it's a covering map), but I can't justify that, and I actually think that it's not true. I think I might need an extra step, and to take an even smaller neighborhood of $y$, in order to make sure that extra sets from $ \lbrace U_i \rbrace $ didn't sneak into $f^{-1}(V)$.  Any help would be greatly appreciated as I've already spent several hours working on this problem.
21296
The more general version of this theorem in Munkres' 'Topology' (p. 290 - 2nd edition) states that   Given a locally compact Hausdorff space $X$ and a metric space $(Y,d)$; a family $\mathcal F$ of continuous functions has compact closure in $\mathcal C (X,Y)$ (topology of compact convergence) if and only if it is equicontinuous under $d$ and the sets  $$ \mathcal F _a = \{f(a) | f \in \mathcal F\} \qquad a \in X$$  have compact closure in $Y$.  Now I do not see why the Hausdorff condition on $X$ should be necessary? Why include it then? Am I maybe even missing something here (and there are counterexamples)?  btw if you are looking up the proof: Hausdorffness is needed for the evaluation map $e: X \times \mathcal C(X,Y) \to Y, \, e(x,f) = f(x)$ to be continuous. But the only thing really used in the proof is the continuity of $e_a: \mathcal C(X,Y) \to Y, \, e_a(f) = f(a)$ for fixed $a \in X$.  Cheers, S.L.
36804
I am studying functional analysis, and in the setting of normed spaces I have seen the theorem that states that the unit ball is compact iff the space is finite dimensional.  I also saw an exercise:     Let $X$ be finite dimensional, prove that for any non-empty and closed   $C$, $x\in C$ there exist $c\in C$ s.t $||x-c||=d(x,C)$.   The proof started with:     take $c\in C$ and consider  $$ D=\overline{B(x,||x-c||)}\cap C $$      $D$ is closed and bounded hence, since $X$ is finite dimensional, is   compact.   I would be glad if someone could help me with the following two questions:  1) It seems that the theorem I read about the compactness also extends to $X$ is finite dimensional iff $\{x,||x-x_{0}||\leq d\}$ is compact for any $x_{0}\in X,d\in\mathbb{R}^{+}$. is this correct ?  2) $C$ is closed, $\overline{B(x,||x-c||)}$ is also closed, hence so is the intersection. since $\overline{B(x,||x-c||)}$ is bounded then so is the intersection.  And so I get that $D$ is closed and bounded. Why can't I conclude that $D$ is compact without the assumption that $X$ is finite dimensional ?
23367
As a part of self study, I am trying to prove the following statement:  Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.   Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
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.
53045
Is there a way to prove this using sequential compactness instead of open cover definitions?  My first gut reaction was that the fact was obvious since we can show that the closed subset $[a,b]$ is compact in $\mathbb R$.