It is well known that in an arbitrary metric space a sequentially compact subset is closed and bounded, under which hypothesis does the converse hold? Or in other words, are there spaces, other than $\mathbb{R}^n$ (or any finite dimensional Banach space) with this property?
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From      A metric space (or topological vector space) is said to have the   Heine–Borel property if every closed and bounded subset is compact.   Any subset of a Euclidean space, including itself, has the Heine–Borel property. I was wondering if there are more general types of metric spaces, topological vector spaces, or whatever space where boundedness and closedness can make sense, such that they also have the Heine–Borel property?   Or does the Heine–Borel property characterize subsets of Euclidean spaces?  Thanks and regards!
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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...".
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$\newcommand{\diam}{\operatorname{Diam}}$ $\newcommand{\R}{\mathbb{R}}$ Consider: D1: Let $(X,d)$ a metric space, we have that $M \subseteq X$ is bounded if $\diam(M) &lt; \infty$ D2: Let $(X,+,\cdot,0,1,&lt;)$ an ordered field, we have that $M \subseteq X$ is bounded if it has an upper and lower bounds. For $M \subseteq R$. Show that $M$ is bounded as a subset of the ordered field $( \R,&gt;)$ if and only if $M$ is bounded as subset of the metric space $(\R,d)$ where $d(x,y) = |x-y|$
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I found the following question about compactness:     A set $S\subseteq \mathbb{R}^n$ has the following property:   if a sequence of $S$ has exactly one accumulation point in $S$, then it converges in $S$. Prove that $S$ is compact.    We know that in $\mathbb{R}^n$ compact sets correspond to closed and bounded sets.   (1) I am trying to prove that $S$ is closed. The condition in the question says if a sequence of $S$ has an accumulation point in $S$ then it converges in $S$. I think instead the condition should be if a sequence of $S$ has an accumulation point( in $\mathbb{R}^n$, not necessarily in $S$ )then it converges in $S$.  Am I correct?  (2) How to prove that $S$ is bounded?  P.S. I have made an edit (there is exactly one accumulation point).  Source of the question:  of this problem set.
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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
62938
I don't have complex analysis at my beck and call, and I only have a low level of knowledge in topology, but I need to prove that this metric space (for any real $r$ and $R$ with $r &lt; R$)$$ X = \{ (x, y) \in \mathbb{R}^2 \ | \ r \leq x^2 + y^2 \leq R \}$$  with the Manhattan metric $d((x_1, y_1), (x_2, y_2)) = |x_1-x_2| + |y_1-y_2|$ is not simply connected.  I've already prooven that it's path connected, and now I need to show there are some points $P$ and $Q$ with two paths between them such that one cannot be continuously 'morphed' into the other.  $ \ $  What I have so far is as follows:  I take $P = (0, r)$ and $Q = (0, -r)$, with $f_0$ being a path from $P$ to $Q$ going clockwise around the circle radius $r$ and $f_1$ being much the same but going counterclockwise.  Now I assume there is a function $g : [0, 1]^2 \to X$ such that:   $g(s, 0) = f_0 (s)$ $g(s, 1) = f_1 (s)$ $g(0, t) = P$ $g(1, t) = Q$   To get the final result, I need to show that this function cannot be continuous, but for the life of me I cannot.  For some context, these are the topics which have been visited during the course, roughly in order of recentness.   Multiple connectedness Simple connectedness Pathwise connectedness Interior points Boundary points Open sets Compactness Complete metric spaces Bounded metric spaces Totally bounded metric spaces Closed sets Closure of a metric space Limit points Cauchy sequences Convergence Continuity Metric space quivalence Metric equivalence   Edit: I might have an argument that works, though it's far from rigourous. We can shrink $R$ to be as close to $r$ as we want, so we can essentially constrain the annulus down to a circle and thus force any path from the left side of the circle to the right side to go through $P$ or $Q$.  So holding $s \in (0, 1)$ constant and varying $t$ must produce a path through $P$ or $Q$ for any $s$. If for some $s$ it passes through $P$ and for some other $s$ it passes through $Q$, then there must be $s_0$ such that $\forall \epsilon &gt;0  \ \exists \delta \leq \epsilon$ st. $s_0$ produces a path through $P$ and $s_0 + \delta$ produces a path through $Q$.  Now we can consider the path $g(s, \frac{1}{2})$, and note that it must have a discontinuity at $s_0$.  Now consider the case where all $s$ produce paths through only one of $P$ or $Q$. WOLOG: $P$. Now, at $t = \frac{1}{2}$, $s$ arbitrarily close to $1$ are mapped away from $Q$, but $1$ is always mapped to $Q$ by definition, so $g(s, \frac{1}{2})$ has a discontinuity at $s=1$. Therefore $g(s, t)$ is not continuous.  This argument is definitely iffy to me, if no one has their own argument (using sufficiently low level concepts), then criticism on the above would be appreciated.
29405
I've got a question: I'm trying to prove that every metric space is compact if and only if the space is sequentially compact. In all the proofs I have found, they used the Bolzano-Weierstrass theorem. Is there a way to prove this fact without using Bolzano-Weierstrass?  Thanks so much!!
50504
Let $A, B$ be subsets of a metric space $X$. If $A$ is compact and $B$ is closed, show that the distance between $A$ and $B$ is achieved.   Attempt at a proof:    Let $A$ be compact and $B$ be closed. Let $m=d(A,B)=\inf_{b\in B} d(A,B)$. Then, there are two possibilities: (a) $\exists b\in B$, $d(a,b)=m$. If this is the case, we're done. (b) $\forall b\in B$, $d(a,b)&gt;m$. In this case,     there exists a sequence $\{b_n\}\subseteq B:$ $d(a,b_n)\rightarrow m$ as $n\rightarrow\infty$ by definition of infinum. Then there exists a subsequence $\{b_{n_k}\}$: $d(b_{n_1},a)&gt;d(b_{n_2},a)&gt;...$ which is monotonic decreasing. Then note that $d(b_{n_k},a)&lt;d(b_{n_1},a)&lt;\infty$. So it's a bounded sequence. Now I want to show that it has a convergent subsequence that converges to $b\in B$ and then I want to do the same for $A$. And then finally to show that $d(a,b)=m$ in fact. Any clues to how to get there?
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Is there an example of a bounded not closed set $A\subset \mathbb R^n$ for some $n\in \mathbb N$ and an isometry $f:A\rightarrow A$ ($\forall x,y\in A\space;\space d(x,y)=d(f(x),f(y))$ where $d$ is the euclidean metric) such that $f$ isn't surjective?
6638
Can you give me an example of a function which is sequentially continuous but not continuous? (I know that in first-countable spaces this is equivalent, but what about in spaces without this condition?)  Thank you :)
70505
Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$
110759
I am given the following problem:     Show that if a metric space (X,d) is compact (meaning X is compact   with respect to the metric d), then there exist points a,b ∈ X such   that d(a,b) = sup{d(x,y) : x,y in X}. (Hint: You’ll need to use the fact   that (X,d) is compact twice. In particular, this means that every   compact metric space has an upper bound on how far away two points can   be)   However, I am not sure if the problem has left out $X \subseteq \mathbb{R}$. Does the statement hold in a general metric space?  Secondly, how might one go about proving this?
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By a d-metric on a set $X$ we mean a function $d : X × X \to \mathbb{R}$ satisfying the following two properties:   $d(x,y)≥0$ for all $x, y∈X $ and $d(x,x)=0$ for all $x, y∈X.$   The $d$-ball of radius $r&gt;0$ centered at $x$ is defined as $B_r(x)=\{y∈X:d(x,y) &lt; r\} $  We'll say the topological space $(X,T)$ is d-metrizable if there is a d-metric $d$ on $X$ such that $U$ is in $T$ if and only if for all $x \in U$ there exists a $d$-ball centered at $x$ which is contained in $U$. (I.e., the family of $d$-balls is a basis for the topology $T$.)  How can I prove that d-metrizable spaces are sequential?  (A topological space $X$ is  means every sequentially open set is open, where $A \subseteq X$ is sequentially open if whenever a sequence $( x_n )_n$ converges to a point in $A$, then $x_n \in A$ for all but finitely many $n$.)
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.
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Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon&gt;0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)&lt;\epsilon$ for $1\leq k\leq n$.  Is the hypothesis that $F$ is closed needed?
37510
If a metrizable topological space is totally bounded with one metric, is it totally bounded with all others? A related, stronger question: if every metrization of a topological space is bounded, are they all totally bounded?
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Let $K$ be a compact metric space. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions on $K$ such that $f_n$ converges to a function $f$ pointwise on $K$.   on Walt. Rudin's book Principles of mathematical analysis, 7.13, if we assume   (1). $f$ is continuous;   (2). $f_n(x)\geq f_{n+1}(x)$ for all $x\in K$ and all $n$;   then it is proved that $f_n$ converges to $f$ uniformly on $K$.   Is there counterexample satisfying (1) but not (2)? And satisfying (2) but not (1)?