How does one find a decreasing sequence of closed balls ( not necessarily concentric) in a complete metric space whose intersection is empty?
121746
Could someone provide me with an example of a metric space having a nested decreasing sequence of bounded closed sets with empty intersection? I first thought of Cantor set but the intersection is not empty!
19650
Let $(X, \rho)$ be a metric space and $\mu^*$ and outer measure on $X$. Then, if every open subset of $X$ is $\mu^*$-measurable, $\mu^*$ is a metric outer measure.   I'm stuck with this problem. I'm supposed to prove that for every $A$, $B$ subsets of $X$ with $d(A, B) = \inf\{d(a,b) \mid a \in A, b \in B \} &gt; 0$,  $$\mu^*(A \cup B) = \mu^*(A) + \mu^*(B)$$ and the hypothesis says that if $U \subset X$ is open, then  $$\mu^*(E) = \mu^*(E\cap U) + \mu^*(E\cap U^c)$$ for every set $E$ but I don't know how to do it. I thought of letting $E = A \cup B$ and then try to find some choice of $U$. With that choice of $E$, we have that  $$ \begin{align*} \mu^*(A\cup B) &amp;= \mu^*((A\cup B) \cap U)+\mu^*((A\cup B)\cap A^c)\\ &amp; =\mu^*((A\cap U) \cup (B \cap U))+ \mu^*((a \cap U^c)\cap (B\cap U^c)) \end{align*}$$  So at first sight, it looks like the open set $U$ we seek for has to be of one of the following two forms, $U \subset A$ and $B \subset U^c$ or the inverse, $U \subset B$ and $A \subset U^c$. The problem is that, even though the existence of an open set $U \subset A$ is always guaranteed, I don't know why it should also satisfy the property that $B \subset  U^c$ as well. My guess is that there is some property of metric spaces that I'm not aware off (or maybe I am really off the track here). Any help would be appreciated. Thank you!
97024
Let $(X,d)$ be a metric space.  Prove:   the union of any open sets in $X$ is open in $X$ the intersection of a finite number of open sets in $X$ is open in $X$   I could prove the first one but how do we prove the second one?
111431
Letting $\mathbb{Q}$ be equipped with the Euclidean metric.  What I can work out is that it is bounded as its contained in the closed ball of radius ${\sqrt2}/{2}$ centred at ${\sqrt2}/{2} $.  Its not compact as it can be expressed as union of the two disjoint open sets $[0,{\sqrt2}/{2}) $and$ ({\sqrt2}/{2}, \sqrt2)$ (though I'm not sure if this makes it not compact in $\mathbb{Q}$ or just $\mathbb{R}$).  And its not closed as the sequence of truncations of $\sqrt2$ in $\mathbb{Q}$ converges to $\sqrt2$
20372
Claim: Let $X$ be a metric space. If $A,B\in X$ are disjoint, if A is compact, and if B is closed, then $\exists \delta&gt;0: |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$.  Proof. Assume the contrary. Let $\alpha_n\in A,\beta_n\in B$ be chosen such that $|\alpha_n-\beta_n|\rightarrow0$ as $n\rightarrow \infty$.  Since A is compact, there exists a convergent subsequence of $\alpha_n\;(n\in\mathbb{N})$, $\alpha_{n_m}\;(m\in\mathbb{N})$, which converges to $\alpha\in A$.  We have   $$|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.$$ Hence $\alpha$ is a limit point of B and since B is closed $\alpha\in B$, contradiction.  Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.
84385
Let $\{K_i\}_{i=1}^{\infty}$ a decreasing sequence of compact and non-empty sets on $\mathbb{R}^n.$ Then $\cap_{i = 1}^{\infty} K_i \neq \emptyset.$  I heard about a proof that take $x_i \in K_i.$ Then I have constructed a sequence. It is bounded because for every $i$, $x_i \in K_1$ that is bounded. Then there is a subsequence that converges. That is where I stopped. What indeed means that a subsequence converges? And why does the limit of such subsequence is in every $K_i,$ showing that such intersection is not empty.  Thank you!
63602
This is not a very well-defined question.  Are there any standard constructions of metric spaces, parameterized by real-valued $n \ge 1$, such that:   When $n$ is an integer, the metric space is precisely $\mathbb{R}^n$. When $n$ is non-integer, the metric space can be seen as a reasonable generalization of $\mathbb{R}^n$.  For example, perhaps it has  of $n$.   Alternately, a non-existence result that you can't maintain some of the important properties of $\mathbb{R}^n$ in a generalization like this would be interesting to me.
59235
This question came up while showing the composition of a metric with a certain other  function gives another metric. Suppose I have some metric space $(X,d)$ and a  continuous, non-decreasing function $f$ on the nonnegative reals. Moreover, suppose  that $f(x)=0$ iff $x=0$, and $f$ also satisfies the triangle inequality in that $f (x+y)\leq f(x)+f(y)$.   Using these properties, it is not difficult to show that $f\circ d$ is yet another  metric on $X$, so $(X,f\circ d)$ is also a metric space.   I notice the (open) balls given by the metrics are of form  $$ B_d(x,r)=\{y\in X\mid d(x,y)\lt r\} $$ and  $$ B_{f\circ d}(x,r)=\{y\in X\mid (f\circ d)(x,y)\lt r\}, $$  so it seems that the topologies generated by the base of open balls in each case  would probably be the same. I would like to see how one would go about showing the topologies given by these two metrics are indeed the same. Thanks for any insight.
91073
Please ,how to prove that the space $\mathbb{R}$ endowed with the metric $d(x,y)=|e^x-e^y|$ is  not a complete space?  I don't find a Cauchy sequence   but not convergent   Please  Thank you.
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.
23792
Let $X$ denote a metric space.  Whenever $S \subseteq X$ and $r \in \mathbb{R}_{\geq 0}$, write $f(S,r)$ for the following set.  $$\{x \in X \mid \exists s \in S : d(x,s) \leq r\}$$     Question. Given a closed set $S \subseteq X$ and any $r \in \mathbb{R}_{\geq 0}$, does it follow that $f(S,r)$ is closed?   If not, a counterexample $(X,S,r)$ would be appreciated.
96247
In a metric space, is it possible to find a sequence which converges to two different limits with respect to two different metrics?  Obviously the metrics can't be equivalent.
80875
Let $X$ be the set of all bounded sequences $x=(x_n)$ of real numbers and let $$d(x,y)=\sup{|x_n-y_n|}.$$ I need to show that $X$ is a complete metric space.  I need to show that all Cauchy sequences are convergent. I appreciate your help.
26289
I have a small question when we have a bouded sequence in a metric space; we say that there exists a closed ball $B'$ such that $(x_n)\subset B'$ or just there exist a ball $B$ such that $(x_n)\subset B$ ?  Thank you.
95191
This question is related to Exercise 2.15 in Baby Rudin, in which the reader is instructed to "[s]how that Theorem 2.36 and its Corollary become false (in $R^1$, for example) if the word 'compact' is replaced by 'closed' or by 'bounded.'"  The relevant theorem and corollary are as follows:  Theorem 2.36: If $\{K_{α}\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{α}\}$ is nonempty, then   $\{K_{α}\}$ is nonempty.  Corollary: If $\{K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n}$ contains $K_{n+1},  (n = 1, 2, 3, ...)$, then   $K_n$ is not empty.  My Question: I realize this question and similar ones have been asked elsewhere on MSE, but I've yet to see a straightforward (to me, at least) answer to my question specifically. Usually, for the "closed" part of the exercise, the following collection of closed sets, $\{K_n \}_{n \in \mathbb{N}}$ is used:  $$K_n = [n, \infty), n \in \mathbb{N}.$$  For, the intersection of every finite subcollection of ${K_n}$ is clearly nonempty. Indeed,  $$ \bigcap_{i=1}^{m} K_{n_{i}} = [N, \infty) $$  where $N = \max\{n_1, n_2, ..., n_m\}$.  Then, the assertion is made that   $$ \bigcap_{n=1}^{\infty} K_n = \emptyset. $$  However, I don't see how the same argument can't simply be used, i.e., that, for example, $N \in  \bigcap_{n=1}^{\infty} K_n$ (or maybe, say, $n+1$). Intuitively, this certainly makes sense, I'd say. For, given the archimedean property, it seems that as $n$ approaches infinity, we can always find an $l$ in each increasingly smaller set such that $n &lt; l &lt; \infty$; and so $l \in  \bigcap_{n=1}^{\infty} K_n$.  What is it exactly that I'm misunderstanding or overlooking here?
61645
Given that $(X,d)$ is a metric space, with $d$ the French railway metric. Show that $(X,d)$ is a complete metric space.  Suppose $\{x_n\}$ is a Cauchy sequence in $(X,d)$. Choose $N$ such that for all $m,n$ $\geq$ N: d($x_{n},x_{m}$) &lt; $\epsilon$. (Note that: d($x_{n},x_{m}$) = d($x_{n},p)+d(p,x_{m}$) , $p \in X$).  Then for all $n \geq N: d(x_{n},x) \leq$ .....  Thats where im stuck. Help would be much appreciated.
14631
Let $(X,d)$ be a metric space where $X$ is an infinite set. Prove that the space has an open set $U$ such that both $U$ and its complement are infinite sets.   I have considered if $d$ is the discrete metric, then one such $U$ exists. But if $d$ is not discrete, how do we find such $U$?