For many of the spaces we work in the closure of open balls are the closed balls, however, I had an example in a class recently that there are some spaces in which this is not true. This is easily proved using either $\mathbb{Z}$ with the euclidean metric or a non-trivial set with the discrete metric.  I was wondering if there is a condition on metric spaces which is equivalent to this intuitive statement being true. I define "the closed balls" as follows:  The closed ball around $z$ of radius $r$, $\hat{B}_r(z) = \{x\in X:d(x,z)\le r\}$  I thought convexity might be a sufficient condition but that's just postulation, don't know how I'd prove it.
27048
It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$ d(x,y)= \begin{cases} 0\qquad&amp;\text{if and only if $x=y$}\\ 1&amp;\text{otherwise} \end{cases} $$ The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything.   I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal?
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!
38249
Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? The closeed-ness comes from Rudin's proof, and the bounded-ness comes from the fact that for a compact subset $E$, we can take the finite open cover and than this open cover is what bounds $E$? Discussions are welcomed! Thank you!!!
73771
A point $p$ in a metric space $X$ is a boundary point of the set $A$, if any neighbourhood of $p$ has points of both $A$ and $X-A$.Prove that the set of all boundary points of $A$ is closed.  My attempt: By definition of an open set this means that for every $x$ in the boundary there is an open ball centred at $x$ contained in the boundary. An open ball is a neighbourhood of $x$, which implies it contains points of $A$ and $X - A$, which in turn implies there are points in both $A$ and $X - A$ that are in the boundary of $A$.  If $A$ is open, then pick any such point in $A$ that is also in the boundary. This point cannot be in $X - A$ by definition of set subtraction. Further, because $A$ is open there exists an open ball centred around this point contained in $A$. Again, an open ball is a neighbourhood, which means a neighborhood of this point does not contain points of $X - A$, implying it cannot be in the boundary, a contradiction. If A is closed then $X - A$ is open and a symmetric argument holds. Hence the boundary is closed.  Is my work correct?
89251
The actual problem looks like-     Let, $(X,d)$ be a metric space where $X\ne\emptyset$. Define $d_1$ and   $d_2$ on $X\times X$ by $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$,   $d_2(x,y)=\min\{1, d(x,y)\}\quad\forall(x,y)\in X\times X.$(It is easy   to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are   equivalent metrics on $X$.   Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent- Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $x\in X$ and each open ball $B_{d_i} (x,r)$ in $(X,d_i)$(for some $r&gt;0$), there is an $s&gt;0$ such that $B_{d_j} (x,s)\subseteq B_{d_i} (x,r)$, for $i, j=1,2$ and $i\ne j$. Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done. I'm able to show $d$ and $d_2$ are equivalent in the following way- Let, $x\in X$ and $B_{d_2} (x,r)$ be an open ball in $(X, d_2)$ for some $r&gt;0$. Then $B_{d} (x,r)\subseteq B_{d_2} (x,r)$. Again, if we take an open ball $B_{d} (x,t)$ in $(X,d)$ for some $t&gt;0$. Then $B_{d_2} (x,s)\subseteq B_{d} (x,t)$, where $s=\min \{1,t\}$.  But how to show $d_1$ and $d$ are equivalent metrics in $X$? Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.
25548
I need to show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric. I know that a set $U$ is open if for $x\in U$ there exists an open ball $B_\epsilon(x)$ such that $B_\epsilon(x)\subset U$ for some $\epsilon &gt;0$. The euclidian metric in $\mathbb{R}^2$ is given by $d_2(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$, so I need to show that for $x\in (a,b)\times (c,d)$ there exists $B_\epsilon(x) = \{y\in\mathbb{R}^2:\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} &lt; \epsilon\}$ such that $B_\epsilon(x)\subset (a,b)\times (c,d)$ for some $\epsilon &gt;0$, but I have no idea how!  Question: How do I show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric?
79824
I am going through Rudin's Principles of Mathematical Analysis in preparation for the masters exam, and I am seeking clarification on a corollary.  Theorem 2.34 states that compact sets in metric spaces are closed. Theorem 2.35 states that closed subsets of compact spaces are compact. As a corollary, Rudin then states that if $L$ is closed and $K$ is compact, then their intersection $L \cap K$ is compact, citing 2.34 and 2.24(b) (intersections of closed sets are closed) to argue that $L \cap K$ is closed, and then using 2.35 to show that $L \cap K$ is compact as a closed subset of a compact set.  Am I correct in believing that this corollary holds for metric spaces, and not in general topological spaces?
50666
Let $X$ be a non empty set and let $C[a,b]$ denote the set of all real or complex valued continuous functions on $X$ with a metric induced by the supremum norm.  How to find open and closed balls in $C[a,b]$? Can we see them geometrically? For example what is an open ball $B(x_0;1)$ i.e. ball centered at $x_0$ with radius $1$ in $C[a,b]$. I can visualize them in $\mathbb R^n$ but when it comes to functional spaces I have no clue how to identify them?   Thanks for helping me.
76402
I'm studying point set topology and looking for a counterexample of "Balls are convex".   We say set $K \subset \mathbb{R}^n $ is convex if $\forall x, y\in K$ implies $\lambda x + \left(1-\lambda y\right)\in K, \forall \lambda \in \mathbb{R}$.   A counterexample has been found in $\mathbb{Q}$ (where the definition of convex has been specialized to "$\lambda\in\mathbb{Q}$") by using p-adic metric. E.g. consider ball $B_{\text{2-adic}}\left(0, 1\right)$.
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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?
85029
I was wondering if this proof of this basic topological result concerning the  closure works.  Proposition: Let $A \subseteq (X,\tau)$. Then, $A$ is dense in $X$ if and only if every non-empty open subset of $X$ intersects $A$ non trivially. Proof: [only if] Assume that $A$ is dense in $X$. This means that $A \cup A' = X$, where $A'$ denotes the set of all the limit points of $A$. Let $G \in \tau$ be non-empty. Thus, $G \subseteq (A \cup A') = X$. Hence, either $G \cap A \neq \varnothing$, which establishes the result, or $G \cap A' \neq \varnothing$, which leads to a contradiction. [if] Assume that $A$ is not dense in $X$. Thus, there is a $x^* \in X \setminus (A \cup A')$. Notice, that $X \setminus (A \cup A')$ is open, because $A \cup A'$ is closed, and it is not empty. Hence, we can conclude by realizing that $X \setminus (A \cup A') \cap A = \varnothing$, which completes the proof. $\square$  It is one of the first proofs I attempt in topology, and I am not planning to see how it is proved in the book I am reading (unless you answer me that my proof is completely wrong). Thus, I am really looking forward to any feedback, also contentwise. Thank you for you time.
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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.
65367
I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
83875
For $x∈\mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $\mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→\mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then which of the followings are true?       $f(B)∩g(B)=\varnothing$  There exist $ϵ&gt;0$ such that $||f(x)-g(x)||&gt; ϵ$ for all $ x∈ B$  There exist $ϵ&gt;0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=\varnothing$ for all $ x∈ B$ ${\rm int }(f(B)) ∩ {\rm int }(g(B))=\varnothing$ , where ${\rm int}(E)$ denotes the  interior of a set $E$      How can I solve this problem? Can anyone help?
112640
Let $X = \{(x,y) \in \mathbb{R^2}: x &gt; 0, y = sin(1/x)\}$. Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?  I just want to make sure I am thinking about these problems in a reasonable manner. I am also trying to apply this to other concepts I have learned.  Is $X$ open?  X is open if every $x \in X$ is an interior point. Take an arbitrary $p \in X$ so that $p = (p_x, p_y)$ then $p_x &gt; 0$ and $p_y = sin(1/p_x)$. Define an open ball $B_r(p)$, and take $q \in B_r(p)$ be an arbitrary point. Is $q \in X$? Is it possible for $q$ to not be in X? Since $y = sin(1/x)$ is defined and is continuous for all $x \neq 0$,then there will always exist a $q_y = sin(1/q_x)$, provided $q_x \neq 0$. How do I know $q_x \neq 0$? Otherwise, I propose that X is open.   Alternatively, X is open if whenever $f: \mathbb{R^2 \to R}$ is continuous and $U \subseteq R$ so that if $U$ is open, then $f^{-1}(U)$ is also open. Could I construct such a $U$ to prove that it's open? My only concern is that if $U = (-1,1)$, then $f^{-1}(U) = (-\infty, \infty) = \mathbb{R}$ which is not open nor closed.   Is $X$ closed?   I am not sure about this one. I want to say that is not closed, since $X^c$ is not open. It is not open because $(0,0)$ defines a boundary point which lies in the set $X^c$.  Thanks!
85905
I want to prove $\Leftarrow$.  I just need one more step, now. If I find a compact set that contains K, then it will finish the proof. Rudin take a k-cell to do that, but before it he proves that every k-cell is compact, and I think I can take a more simple compact to do this step. I has thinked in the closed ball, but don't know how to prove that it is compact. Can anyone help me to find this set, if it really exists? Thanks.
9008
Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete.  My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not figure out the exit way. Maybe I am on the wrong track. Could you please help me?  edit: More from my attempt:  Suppose $A$ is a closed set and let $(x_n)$ be a sequence of points $A$ such that $\lim_n x_n\to b$.  Suppose now that $A$ has the property that $b\in A$, whenever $x_n$ converges to $b$. We know that every element of $x_n$ which is convergent in $X$ also converges to a point in $A$. Since $x_n$ is a Cauchy sequence in $A$, it must converge to a point $y\in A$. But the limit of a convergent sequence is unique.   Take $x\in A$ and select an appropriate $n$, which enables $x_n$ to converge to a point $x$ in $X$. Since the limit is unique, it must follow that $x=y$. Thus $x\in A$ and $A$ is closed.   If $A$ is a closed subset of $X$, then any Cauchy sequence of a point in $A$ is convergent in $X$ and hence converges to a point in $A$. Thus $A$ is complete.