Let $ (X,d)$ be a metric space and $A \subset X$. How can I show that $(A^\circ)^\circ = A^\circ $? $A^\circ$ denotes the set of all interior points of $A$.    I know that $x \in A$ is interior if there exists an $ r&gt;0$ such that $B_r(x) \subset A$ .  So i think to try to prove that $(A ^\circ)^\circ \subset A^\circ$ and $A^\circ \subset (A^\circ)^\circ$ , but I cannot think how to start.
122289
Say we want to show that the interior of a set $A$ is open. If $x \in Int(A)$, then there exists an open ball $B_r(x) \subseteq A$. Since $B_r(x)$ is open, $y \in B_r(x)$ also has an open ball $B_s(y) \subseteq B_r(x) \subseteq A$, so $y \in Int(A)$. Now, somehow we have to show that the ball $B_r(x) \subseteq Int(A)$, and that would complete the proof. All of the proofs I read say this is obvious, but I don't see how $B_r(x) \subseteq Int(A)$ immediately follows here.
80122
I have troubles showing the following:  Let $(X,\rho)$ be a compact metric space and $F \subset X$ a closed subset. Prove that if diam $F &lt; \infty$, then there exist $x_{0}, y_{0} \in F$ such that diam $F= \rho(x_{0},y_{0})$.  It looks so trivial yet hard to find the trick.
18771
Let $\mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $\bigcap\mathscr{F}\ne\emptyset$. Prove that $\bigcup\mathscr{F}$ is connected.   If $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $\bigcap\mathscr{F}$. Then either $x\in A$ or $x\in B$. I don't know where to go from here.
70067
Let $B$ be a compact subset of a metric space $X$. Let $x \in X \setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x \in U$ and $B \subseteq V$.   Is this an existence proof? If yes, do I get to define $U,V$?
6038
Let $X$ and $Y$ be metric spaces, $X$ compact, and $T:X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism.   My attempt:  We need only show that $T^{-1}$ is continuous.  Let $M \subset Y$ be closed. Then, from the surjectiveness of $T$, there exists a $A \subset X$ such that $T(A)=M$. Since $M$ is closed, by the continuity of $T$, we must have that $A$ is closed.  Now $$T^{-1}(T(A))=T^{-1}(M)$$ i.e, $A=T^{-1}(M)$. Since $A$ and $M$ are closed, we must have that $T^{-1}$ is continuous.  My problem with this, however, is that I am not using the fact that $X$ is compact anywhere in the proof. Is there perhaps another way that I can show this, using the compactness of $X$?   Any hints will be appreciated.
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.
77010
let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is closed.       Can a finite subset of $X$ be open ?     Definitions:   a set $F\subseteq X$ is closed (in$(X,d)$) if  $\bar F =F$. a set $U\subseteq X$ is open (in$(X,d)$) if  $U^o=U$
32597
Let $(X,d)$ be a metric space and $A\subset X$ and $x\in X$. Then  $x\to d(x,A)$ is a uniformly continuous function.  If $\partial A=\{x\in X\,:\,d(x,A)=0\}\cap\{x\in X\,:\,d(x,X-A)=0\}$, then $\partial A$ is closed for any $A\subset X$.  If $A,B$ are subsets of $X$  then $d(A, B)=d(B,A)$. If the function is uniformly continuous then $d(x,y)&lt;\delta $ implies $d(d(x,A)-d(y,A))&lt;\epsilon$ , I can not handle the last expression. Difficulty continues for 3rd choice also. At least give me some hints. And for the 2nd choice I think it is true, as it is intersection of two closed sets. As finite intersection of closed sets closed $\partial A$ closed. And logic for former two sets closed is that the sets are preimage of closed set (singleton $\{0\}$ is closed as $\mathbb R$ is $T_1$) under continuous map. I am not very much sure about my ideas and want a verification.
59730
Prove: The set of interior points of any set $A$, written int($A$), is an open set.  Let $p\in$ int($A$), then by definition $p$ must belong to some open interval $S_{p}\subset A$. Now since we know that the real line itself is open then $S_{p}\subset \mathbb{R}$. Now suppose we pick some $q\in A$. I want $q\in$ int($A$).   I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.    Definition of interior: Let $A$ be a set of real numbers. A point $p\in A$ is an interior point iff $p$ belongs to some open interval $S_{p}$ which is contained in A: $p\in S_{p}\subset A$  Definition of open: A set $A$ is open iff each of its points is an interior point
67208
Let $(X,d)$ be a metric space and $ a \in X $ Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$ Use this fact to show $S$ is an open subset of $X$: Let $(X,d)$ be a metric space and $a \in X, r &gt; 0.$ Let $S$={$x \in X : d(a,x) &gt; r$}  &gt;Lipschitz Defn:  Let $f$ be a function from $(X,d)$ to $(Y,\rho)$ $f$ is said to be Lipschitz if: $\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$  &gt;Continuity Defn:  Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$ Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B $} is an open subset of $X$, whenever $B$ is an open subset of $Y$  &gt;Triangle Inequality for a Metric:  $d(x,z) \leq d(x,y) + d(y,z)$   My first question is, am I thinking about this correctly? If $f(x)$ is a Lipschitz function, it is uniformly continuous If $f(x)$ is uniformly continuous, it is continuous at every point If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open Hence $S$ is an open subset of $X$.  Secondly, how would you formally construct this? We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$ By definition of Lipschitz we have $\exists  K \geq 0$  s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$ &amp; since our codomain is $\Re$ $\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$ Taking $K=1$ $\ldots$ Help in finishing this off would be appreciated. Thanks.
18725
Let $X$ be a compact metric space and $Y$ any metric space.  If $f:X \to Y$ is continuous, then $f(X)$ is compact (that is, continuous functions carry compact sets into compact sets). Proof:  Consider an open cover of $f(X)$. Then $f(X) \subset \bigcup_{\alpha \in A}V_\alpha$ where each $V_\alpha$ is open in $Y$. $X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)$. Hence $\bigcup_{\alpha \in A}f^{-1}(V_\alpha)$ is an open cover of $X$.  Since $X$ is compact then we can choose a finite subcover $\{V_i\}_{i=1}^n$ such that $X \subset \bigcup_{i=1}^n f^{-1}(V_i)$. So then $f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i$, a finite subcover of $f(X)$.  $\therefore f(X)$ is compact.  Does this proof have an error? Thanks for your help.
90330
I'm teaching my self topology using a book I found. This is the second part of a 4 part question. First part is .  I'm trying to prove the following problem from a book I found:     Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}$   *where the over line indicates closure.  First I'm going to show an example the them being equal, and then show an example of a proper superset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this.     (I) Example of $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} = \bigcup \limits_{A \in \mathscr{A}} \overline{A}$   I find that with "most" collections of subsets, the two sides come up equal. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=\{(1,6),(3,10)\}$. So then $ \bigcup \limits_{A \in \mathscr{A}} A= (1,10)$, and  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10]$. Next the right side would be: $\bigcup \limits_{A \in \mathscr{A}} \overline{A}= [1,6] \cup [3,10]=[1,10]$. So it is possible for the two sides to be equal.     (II) Example of $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} \supset \bigcup \limits_{A \in \mathscr{A}} \overline{A}$   The only way I know of involves a collection of subsets that "expands" in a way that converges - ie  $\bigcup_{n=1}^{\infty} \overline {B_{\frac{R \cdot n}{n+1}}}(x)=B_{R}(x)$. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=(1,\frac{10n}{n+1}), n\in\mathbb N$. So then $ \bigcup \limits_{A \in \mathscr{A}} A= (1,10)$, and  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10]$.   Next the right side would be: $\bigcup \limits_{A \in \mathscr{A}} \overline{A}= \lim \limits_{n \rightarrow \infty} [1, \frac{10n}{n+1}]= [1,10)$ . And $[1,10] \supset [1,10)$. So it is possible for Closure of Unions to contain Union of Closures.     Proof: $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}$   (II) illustrates that the infinite union of closed sets can make an open (or half open) set. So: * $\bigcup \limits_{A \in \mathscr{A}} \overline{A}$ can be (half)open.  * $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}$ must be closed. Since we're dealing with the same collection of subsets, then the difference will be at the boundary points. And thus, the closed set will contain the (half)open set.  QED  The part of the proof i'm not sure about is at the end where I say, "the difference will be at the boundary points." In (II) that is the case, but will that always be the case when dealing with the most general topological spaces?
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?
115957
Let $A\subseteq X$. If $d$ is a metric for the topology of $X$, show that $d\restriction_{A\times A}$ is a metric for the subspace topology on $A$.   I've shown that $d'=d\restriction_{A\times A}$ is a metric on $A$.  I am letting $\tau_B$ denote the subspace topology on $A$ that is induced by the metric topology on $X$ and $\tau_{B'}$ denote the topology on $A$ induced by the metric $d'$. My goal is to show $\tau_B=\tau_{B'}$.  Let $B=\{A \cap B_\varepsilon^d(x) \mid x \in X, \varepsilon&gt;0\}$, a basis for $\tau_B$ and $B'=\{B_\varepsilon^{d'}(a) \mid a \in A\}$, a basis for $\tau_{B'}$. I've shown $B' \subseteq B$ and so $\tau_{B'} \subseteq \tau_B$.  For showing $\tau_B \subseteq \tau_{B'}$ I've picked some $a \in A$ and $A\cap B_\varepsilon^d(x)$, a basis element of $\tau_B$ such that $a \in A\cap B_\varepsilon^d(x)$. My goal is to find some basis element $C$ of $\tau_{B'}$ such that $a \in C \subseteq A\cap B_\varepsilon^d(x).$ If $x \in A$, then $$A\cap B_\varepsilon^d(x)=B_\varepsilon^{d'}(x)$$ and so I have my needed basis element $C$. The part where I have been having trouble is if $x \notin A$. In this case, what I have tried so far is to see if I can find some $\delta &gt;0$ such that $$a \in B_\delta^{d'}(a) \subseteq A\cap B_\varepsilon^d(x)$$ but in searching for a delta and trying to show via set containment that the $\delta$-ball is contained in the $\varepsilon$-ball intersected with $A$ has been difficult. I've tried using triangle inequality but I am running into a problem of not knowing how to show that if $y \in B_\delta^{d'}(a)$, meaning $d'(a,y)$ that $y \in A\cap B_\varepsilon^d(x)$, mainly that $d(x,y) &lt; \varepsilon$. It seems with the different $\delta$'s I have tried, I can use the triangle inequality to show something like $d(x,y) &lt; \frac{3}{2}\varepsilon$ but not quite $\varepsilon$.
5443
Let $X$ be a metric space with metric $d$. Define $d: X \times X \to \mathbb{R}$, show that $d$ is continuous.  I would like to show that the function is continuous the topology way (since it is a course on topology).   Let $(a,b)$ be a basic open set in $\mathbb{R}$, then $d^{-1}(a,b) = \{(x,y): a&lt;d(x,y)&lt;b\}$. Define $A= \bigcup_{x \in X} B_{b} (x)$, where $B_b (x) = \{y \in X| d(x,y)&lt;b$}. Clearly $A$ is open. Now define $C = \{(x,y):d(x,y)&gt;a \}.$ Essentially I hope that $A \cap C$ is open. But I am stuck in showing that $C$ is open.  Any help please? :)
44318
The question is is the interior of closure of an open set equal the interior of the set?  That is, is this true:  $(\overline{E})^\circ=E^\circ$  ($E$ open)  Thanks.
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!