If $X$ locally compact Hausdorff space. Then  $C_{c}(X)$ is complete implies $X$ is compact.  I know that $C_{c}(X)$ dense in $C_{0}(X)$. So in that case  $C_{c}(X)=C_{0}(X)$. I know only Tiez extension theorem on locally compact Hausdorff space and vary basic knowledge of Banach Algebra.
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.
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.
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.
94054
I have seen two definitions of locally compact topological space   A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists a compact set that contains an open neighborhood of $x$. A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists an open neighborhood of $x$ whose closure is compact.   Are these two definitions equivalent? It's clear that the second implies the first, but I don't see why the first implies the second. Suppose that $U $ is an open neighborhood of $x$ with $x\in U\subset K$, here $K$ is a compact set. Since compact sets may not be closed in a non-Hasudorff topological space, the closure of $U$ may not be contained $K$, so we can't say the closure of $U$ is compact.
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!
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 $X$ ?
79590
Given a compact Hausdorff space $X$, if $a \in X$ is a $G_\delta$ singleton (i.e., $\{ a \}$ is a $G_\delta$ set), then there is a countable local base at $a$.   The point $a$ can be written as countable intersection of open sets let's call $U_{n}$.  How can I use the compactness? What is the next step for proof? Could you give me hint please?
9621
If $(X,d)$ is limit point compact, show it is compact.  I have found multiple proofs of this that first show that limit point compact implies sequential compact, which in turn implies compactness. I was wondering if there is a direct proof showing that limit point compactness implies compactness in a metric space.
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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.
81241
I need to proof that $(C_0([0,\infty)), \ \|\cdot\|_\infty)$ is complete. I think that I did it correctly (or atleast am thinking in the right direction), but I still struggle with writing it down correctly. I used that $(C_b([0,\infty)), \|\cdot\|_\infty)$ is complete, so therefore as $C_0[0,\infty) \subset C_b[0,\infty) $, I wanted to use that if $A$ is a closed subset of a complete space $(M,d)$, that $(A,d)$ is complete. Thus, it came down to showing that $C_0[0,\infty)$ is closed.  Let $f_n(k)$ be a sequence of elements out of $C_0[0,\infty)$ that converges to an element $f$ out of $C_b[0,\infty)$. I need to proof that $f(k) \in C_0[0,\infty)$.   Per the definition of convergence, we know that $\|f_n(k) - f(k) \|_\infty \rightarrow 0$ whenever $n \rightarrow \infty$. We use this to find $ |f_n(k)-f(k)| \leq\|f_n(k) - f(k)\|_\infty  \rightarrow 0$.  Then we consider $|f(k)| = |f(k) - f_n(k) +f_n(k)| \leq |f(k)-f_n(k)| + |f_n(k)|$. If we take limits on both sides, we get: $\lim_{k\to \infty} |f(k)| \leq \lim_{k\to \infty} |f(k)-f_n(k)| + \lim_{k\to \infty} |f_n(k)| = 0 + 0 = 0$ (the first term follows from what we have previously stated and the second term follows from the fact that $f_n(k)$ is an element of $C_0$. Thus, $f \in C_0$ and therefore $C_0$ is closed and therefore it is complete.  My questions are the following;   What exactly does it mean to be a sequence of functions? I used $f_n(k)$ and $f(k)$ in my proof because I saw that it was used in another exercise, but I'm not sure what it means to write that $k$ down. Is the structure of my proof sound? Is there a better way to proof such statements?   Thanks for reading,  K. Kamal
60535
Let $C$ be the Banach space of all complex continuous functions on $[0,1]$ endowed with the supremum norm. Let $B$ be the closed unit ball of $C$. Does there exist a $\Lambda\in C^*$ for which $\Lambda (B)$ is open? Note that $C$ is a locally convex space and $B$ is convex, balanced and closed set, by the separation theorem of convex sets,we know there exists $\Lambda\in C^*$ such that $|\Lambda f|\leq 1$ for all $f\in B$ and $|\Lambda f|&gt;1$ for $f\in B^c$. But how about this when we consider some concrete space above?
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 $Y$?
27658
Is it possible to have a topological space having all compact subsets closed, but the space itselt is not Hausdorff?  I couldn't find any counterexample. I tried $\mathbb R$ with cofinite, point inclusion and point exclusion topologies, but these didn't work.  Any suggestion in this context will be helpful for me.
28852
While studying measure theory I came across the following fact: $\mathcal{K}(X) \subset C_b(X)$ (meaning the continuous functions with compact support are a subset of the bounded continuous functions). This seems somehow odd to me; I've tried to prove it but did not succeed. Could someone help me out here?  Thanks!
109823
Let $X$ be a Hausdorff space and let $C_0 \supset C_1 \supset ...$ be a decreasing sequence of compact connected subsets of $X$.  Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected.  I have the compact part.  Just don't know how to show it is connected.
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.
99860
Show that a totally bounded complete metric space $X$ is compact.   I can use the fact that sequentially compact $\Leftrightarrow$ compact.   Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$ $\forall\epsilon&gt;0$, $X$ can be covered by a finite number of balls of radius $\epsilon$. I'm trying to show that all sequences in $X$ have a subsequence that converges to an element in $X$. I don't see how to go from convergent Cauchy sequences and totally bounded to subsequence convergent $in$ $X$.