$X$ is a compact metric space. Let $A \subseteq X$ be a compact set. Let $\{C_n\} \in A$ be a Cauchy sequence. Because it is Cauchy, $\{C_n\}$ must converge somewhere (although it doesn't need to converge in $A$). Let the limit point for $\{C_n\}$ be denoted by $C$. We also have that $A$ is compact $\implies A$ is closed and bounded $\implies$ $C \in A$ (because closed sets include the limit points for all its convergent sequences) $\implies$ A is complete.
77996
I need to prove that every compact metric space is complete. I think I need to use the following two facts:    A set $K$ is compact if and only if every collection $\mathcal{F}$ of closed subsets with finite intersection property has $\bigcap\{F:F\in\mathcal{F}\}\neq\emptyset$. A metric space $(X,d)$ is complete if and only if for any sequence $\{F_n\}$ of non-empty closed sets with $F_1\supset F_2\supset\cdots$ and $\text{diam}~F_n\rightarrow0$, $\bigcap_{n=1}^{\infty}F_n$ contains a single point.   I do not know how to arrive at my result that every compact metric space is complete. Any help?   Thanks in advance.
27054
I was given the following question: Let $q$ be a real number such that $0 &lt; q &lt; 1$, and let ${a_n}$be a sequence that follows the rule : $|a_{n+1}-a_n| &lt; q|a_n-a_{n-1}|$ for any $n&gt;1$. Prove that ${a_n}$ converges. I already know this can be proven using Cauchy's criterion, but I had a different proof in mind (using Cantor's lemma), and wanted to know if it's valid, or if I might be mistaken. So here it is: First - I prove that the sequence is bounded. Since the sequence is bounded, the set of its members (and the set of any sub sequence of ${a_n}$) has a Supremum and an Infimum. Then, for any natural number $k$, I define the closed interval $A_{k}$, such that $A_{n}=[\inf(a_k), \sup(a_k)]$, where inf and sup of $k$ are the Infimum and Supremum of the set of the members of $a_n$, starting from $a_k$ (meaning, any member with an index greater than or equal to $k$). Naturally, $A_1$ is the &quot;biggest&quot; interval, and because of how the sequence is defined, $A_{n+1}$ is always a proper subset of $A_n$ for any $n$. Now I use Cantor's lemma - because the interval gets infinitely smaller and smaller (after n iterations, it gets at least $q^n$ times smaller), the intersection of all intervals $A$ (meaning, $A$ of $n=\infty$) contains one single point, let's call it $L$. Now, I state that $L$ must be the limit of $a_n$ - for any epsilon, there exists a natural number $k$, such that for any $n&gt;k$, $A_k$ is contained in the epsilon neighborhood of$ L$, and therefore all members of $a_n$ starting from $a_k$ are in the epsilon neighborhood of $L$, which means $L$ must be the limit. So do you think this is a valid proof? Or am I making some false assumptions? I know this proof is a bit longer than using Cauchy's criterion, but I like it better because I feel like it gives a better explanation as to 'why' the series converges. Thanks in advance!
121620
In metric space,  $(X,d)$  I know limit points of some set(S maybe) is closed in X  Then is set of isolated points of S is closed?   Want to make sure if closure of isolated points of S doesn't intersect with limit points of S  But i think i got it and i think i should leave it for people like me
33925
Consider the metric space $Q$ of rational numbers with the Euclidean metric of $R$. Let $S$ consist of all rational numbers in the open interval ($a, b$), where $a$ and $b$ are irrational. Then $S$ is a closed and bounded subset of $Q$ which is not compact.  I am trying to prove this statement. It's easy to show that $S$ is closed and bounded but I'm having trouble showing that it is not compact. I think I need to find an example of an open covering of $S$ which doesn't have any finite subcover, but I can't think of such an example. Any help? Thanks.
3946
$X$ is a compact metric space, $C(X)$ is separable when $X$  is compact where $C(X)$ denotes the space of continuous functions on $X$. How to prove it?  And if $X$ is just a compact Hausdorff space, then $C(X)$ is still separable? Or if $X$ is just a compact (not necessarily Hausdorff) space, then $C(X)$ is still separable?  Please help me. Thanks in advance.
65179
Can someone please check whether the following simple proof is "mathematical"? Is it correct, complete, rigid? Can it be simplified? I'm a complete autodidact so I'm looking for someone to give me feedback to gain experience in writing proofs... This is also my first question on MSE.  The proposition:  Let $(X, d)$ be a metric space and $x_n \to x$ where each $x_n \in X$ and $x \in X$. Let $A$ be the subset of $X$ which consists of $x$ and all of the points $x_n$. Prove that $A$ is closed in $(X, d)$.  My tentative to prove this:  We first show that all infinite sequences in $A$ converge to $x$: Let $y \in X$, $y \ne x$. Then there is some open ball $B_\epsilon(x)$ with $\epsilon &lt; d(x,y)$ containing all but finitely many elements of $A$. As $y \notin B_\epsilon(x)$ there can be no infinite sequence in $A$ converging to $y$. Consequently all infinite series in $A$ converge to a point in $A$ which therefore must be a closed set.  Edited: As rightly pointed out in the comments, I should have written in the first sentence "...sequences with infinitely many distinct terms and which converge to some point of $X$" and the last sentence should be "Consequently all infinite sequences...".
18101
Let $(X, d)$ be a metric space. I have to prove the following statements are equivalent.   $(X,d)$ is complete (i.e., every Cauchy sequence is convergent) and totally bounded (i.e., for every $\epsilon&gt;0$, $(X,d)$ has a finite $\epsilon$-net). $(X,d)$ is sequentially compact (i.e., every sequence has a convergent subsequence) . $(X,d)$ is Bolzano - Weierstrass compact (i.e., every infinite subset of $X$ has a limit point). $(X,d)$ is compact (i.e., every open cover of $X$ has a finite subcover). Every infinite open cover of $(X,d)$ has a proper subcover.   I have proved the following: 1$\Longleftrightarrow$2, 2$\Longleftrightarrow$3, 4$\implies$3, 2$\implies$4, 4$\implies$5 and 5$\implies$3. Although they are enough to prove the equivalence of all five statements, I tried to prove it from the other possible directions. But I'm stuck with the proof of 3$\implies$4 and 5$\implies$4. Does anybody know how to prove them without using any of the above statements? Please help.
94742
I found at the beginning of tom Dieck's Book the following (non proved) result     Suppose $X$ is the colimit of the sequence $$ X_1 \subset X_2 \subset X_3 \subset \cdots $$ Suppose points in $X_i$ are closed. Then each compact subset $K$ of $X$ is contained in some $X_k$   Now I really don't know how to prove this fact. The idea would be to find a suitable open cover to it and after taking a finite sub cover trying to claim that $K$ lies in one of the $X_k$. I'm able to do this reasoning in some more specific cases, where I've more control on how open subsets looks like, but in this full generality I don't see which open cover I can take.     My Attempt: The only idea or approach I'm able to cook up so far is to try use some kind of sequence of points $x_n \in K\cap X_n \setminus X_{n-1}$ which can be assumed to exist by absurd. Being $K$ compact, there must be an accumulation point $k\in K$. Clearly $k \in X_k$ (little abuse of notation here) and for every neighbourhood of $k$, there is a tail of this sequence entirely contained in it. Now everything seems to boil down to find the right nbhd to find the counterexample. It seems doable, but I don't have any idea on how to choose it, because the only open I have for sure are complements of points, but they seems a little bit coarse for what I want to do.   As a side note, May claims at page $67$ of his "Concise Course (revised)" that this result holds for any based spaces. The proof seems to use the above result without T1 assumption. How one can prove this result in such generalities? (no details where provided, only the rough idea.
19317
This is the analogue of closed graph theorem for compact space          Suppose that $X$ and $K$ are metric spaces, that $K$ is compact, and that the graph of $f: X \rightarrow K$ is a closed subset of $X \times K$. Prove that $f$ is continuous.      Because metric space is first countable, all we need is to prove that if $x = \lim_{n \rightarrow \infty} x_{n}$, then $f(x) = \lim_{n \rightarrow \infty} f(x_{n})$. Suppose $x = \lim_{n \rightarrow \infty} x_{n}$, then because $K$ is compact, sequence $\{f(x_{n})\}$ has a subsequence $\{f(x_{n_{i}})\}$ converge. Because the graph of $f$ is closed, we must have $\lim_{n \rightarrow \infty}\{f(x_{n_{i}})\} = f(x)$. All we need to prove now is $\{f(x_{n})\}$ converge to $f(x)$, but I can't prove that. Can any one help me. I really appreciate. Thanks
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.
2658
I know that this question has been asked to death, and multiple solutions are given, but I still don't understand why the "standard" proof works  Following     Let $C_o([0,1])$ be the space of continuous functions on $[0,1]$. Define $\|f\| = \sup\{|f(x)| x \in [0,1]]\}$  Let $B[0,1] = \{f \in C_o|\|f\|\leq 1\}$, show that this Closed, Unit ball in $C_o$ is not compact.  Proof Sketch:     Let $\{f_n\} \subset C_o, f_n = x^n, x \in [0,1]$, then $f_n \in C_o,  \|f_n\| = 1 \thinspace \forall n$, so $f_n \in B([0,1]) \thinspace \forall n$      But $f_n \to f \notin C_o$. So $B([0,1])$ is not compact.   Three questions:    if $f_n \to f \notin C_o$, wouldn't that mean $B([0,1])$ is not closed in the first place (contradicts with assumption)? How does the above implies that no subsequence converges in $B([0,1])$ Is the proof sketch essentially correct?
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.
30915
I'm sure I'm having a notational misunderstanding. Anyway, suppose $(f_n)$ is a sequence of continuous functions from a metric space $X$ to $\mathbb{R}$. So, if $(f_n)$ converges uniformly to a function $f$, then  $$\lim_{n \to \infty} \inf_{x \in X} f_n(x) = \inf_{x \in X} \lim_{n \to \infty} f_n(x)$$  and  $$\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x).$$  Can somebody explain to me why is that? Again, this might be really simple...
22975
I found the following property of compact operators in a proof, and I can't prove it.     Prove that if $T \in \mathcal{L}(E,F)$ is compact, and if $u_n \rightharpoonup u$ (the sequence converges weakly to $u$ in $\sigma(E,E^*)$) then $Tu_n \to Tu$ strongly in $F$.   I was able to prove that $Tu_n$ has a convergent subsequence ($u_n \rightharpoonup u$ implies that $(u_n)$ is bounded in $E$. Then, because $T$ is compact it must follow that $(Tu_n)$ must contain some strong convergent subsequence), but didn't manage to finalize the proof.  Any reference or hints are welcome.
14517
I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete.  The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in \mathbb{R}$.  I don't understand why it's not enough that $f$ is bounded. I think the limit of a sequence of continuous functions is continuous and then if $f$ is bounded, it's in $C_b(X)$. So what is this $\lim f_n \in \mathbb{R}$ about? Many thanks for your help.
24663
A set $A \subset \ell_1$ is compact if and only if $A$ is closed and bounded and given any $\epsilon &gt;0$, there exists $n_0$ such that $\sum_{k=n}^{\infty} |x_k| &lt; \epsilon$ for all $n&gt; n_0$ and $x \in A$.   It's easy to prove the part assuming that $A \subset l_1$ is compact, but I am finding difficulty to prove the other part.
8222
A‎‎ ‎sequential ‎space ‎has ‎unique ‎sequential ‎limits ‎iff ‎each ‎countably ‎compact ‎subset ‎is ‎closed.  ‎Proof: ‎If‎ ‎$ \{ x_n \} $ ‎is a‎ ‎sequence ‎converging ‎to ‎two ‎distinc ‎‎$‎x‎$ ‎and ‎‎$‎y‎$‎, then ‎$ ‎\{ x‎ ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n  :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎$ ‎is a‎ ‎non ‎compact ‎set.  conversely: ‎if ‎‎$‎X‎$ ‎is ‎sequential ‎and ‎has ‎unique  ‎sequential ‎limits, ‎‎$‎X$ ‎is ‎‎$‎T_1‎$ .‎ ‎Let‎ $ K‎ ‎‎\subseteq ‎X $‎ ‎be ‎countably ‎compact.If ‎‎$ ‎\{ ‎x_n ‎\} ‎‎$ ‎is a‎ ‎sequence ‎in ‎‎$‎K‎$ ‎and‎ $ ‎\{‎x‎_{‎n}‎  : n‎ \in‎ \omega ‎\} ‎‎\longrightarrow ‎x‎_{0}‎$ ‎then‎  ‎$ ‎\{ ‎x‎_{0} ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n  :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎$ ‎is ‎sequentially ‎closed ‎and ‎hence ‎closed. ‎Thus ‎‎$‎x_{0}‎$ ‎is ‎the ‎only ‎possible ‎accumulation ‎poin ‎of‎ $ ‎\{ ‎x‎_{‎n}‎  : n‎ \in‎ \omega ‎\} ‎‎$‎. if $ ‎\{ ‎x‎_{‎n}‎  : n‎ \in‎ \omega ‎\} ‎‎$ ‎is ‎infinite‎, ‎$‎x_0 ‎\in ‎K‎$‎. ‎If ‎not ‎for ‎some ‎‎$‎n‎$‎, $ ‎x‎_{0} =‎ ‎x‎_{n}‎‎ ‎\in K ‎$.Hence ‎‎$‎K‎$ ‎is ‎closed.‎     (1) Why" ‎if ‎‎$‎X‎$ ‎is ‎sequential ‎and ‎has ‎unique  ‎sequential ‎limits, ‎‎$‎X$ ‎is ‎‎$‎T_1‎$?      (2) Why "if $ ‎\{ ‎x‎_{‎n}‎  : n‎ \in‎ \omega ‎\} ‎‎$ ‎is ‎infinite‎"?