Prove the set of limit points in a $T_1$ space is closed.  Definitions:  Limit Point: $x$ is a limit point of a set if every open set that contains $x$ contains at least one other point of the set that is not $x$. $T_1$-space: All pairs of disjoint points have neighborhoods not containing each others' points.  I wrote that all points in the set of all limit points are disjoint from the complement of the set of all limit points, so all the points in the complement of the set of all limit points have neighborhoods that are disjoint from all the neighborhoods of all the points in the set of all limit points, but this was marked incorrect. How should I go about proving this correctly?
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Let $X$ be a Hausdorff space and $A\subset X$.  Define $A'=\{x\in X\mid x\text{ is a limit point of }A\}$.  Prove that $A'$ is closed in $X$.  Relevant information:  (1.) Every neighborhood of a point $x\in A'$ contains a point $y\in A'$ distinct from $x$ (in fact, in a Hausdorff space, every neighborhood of $x$ contains $\infty$-many points of $A$ distinct from $x$)  (2.) In a Hausdorff space, every sequence has a unique limit.  On first glance, it should be an easy proof, but I've made little progress.  I was planning on showing $\overline{A'}=A'$.  Firstly, $A'\subset \overline{A'}$ trivially.  To show $\overline{A'}\subset A'$, proceed by contradiction.  Assume there exists $x\in\overline{A'}$ such that $x\not\in A'$.  This should yield an easy contradiction but I don't see it.  In particular, I'm unsure if the fact that $x\in\overline{A'}$ implies that there actually exists a sequence in $A'$ converging to $x$.  By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences?  If this is true, it should yield an easy contradiction.  Any help?
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It just seems like I can't wrap my head around this. For a set $A$ how can I show that the supremum of that set $\sup(A)$ is also a limit point of said set? Can I just set $$a=\sup(A)$$ $$b\le a \: \:  \forall \: b \in A$$ and then show that there's a point different from $a$ in any $\epsilon$-neighbourhood for any $\epsilon&gt;0$? Or should I use a completely different approach?
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I have a non homework related question from a text and require a nice clear proof/disproof please  Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?  my ideas:  if $H$ is the subset of the topological space $X$  if the subset is closed in the closed subspace, the complement is open in the subspace, which means the complement is of form $U\cap H$ for some $U$ open in $X$  if the subspace is closed the complement is open which means complement of $H=U$ for some open $U$ in $X$  kind thanks
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Theorem:   Let $S_0, S_1 \subseteq \mathbb{R}^n$ be disjoint and closed.   Then there exist disjoint open sets   $T_0, T_1 \subseteq \mathbb{R}^n$ such that   $S_0 \subset T_0, S_1 \subset T_1$.   How to prove this theorem without Urysohn functions? I know that $S_0^c$ contains $S_1$ and vise versa, but I don't know how to proceed from here.
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The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$  $X$ and $Y$ are metric spaces.  a) Suppose $f$ is continuous and prove that $G(f)$ is a closed set.  b) Suppose that $G(f)$ is compact and prove that $f$ is continous  For a), the definition of a closed set that comes to my mind is a set that contains all its limit points (or was it accumulation points?), is there another equivalent definition that may b more helpful to prove a)? Is it possible to prove this directly? Because at first glance the only way I could imagine to prove this is by contradiction or contrapositive.   I imagine that the proof of b) will be immediately derived from a).
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My instructor proves a statement during a lecture: if a set $E$ is open in a metric space $X$, then $E^c$ is closed.  In his proof, he writes: Suppose $E$ is open. We want to show that $E^c$ is closed (contains all its limit points). Let $x$ be a limit point of $E^c$. Then......  My question is how to know there exists a limit point in $E^c$? Thanks!
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The closed unit interval $\mathbb{I}=[0,1]$ is a connected subset of   $\mathbb{R}$.   I am having difficulty understanding the proof in my book, which goes:     Suppose that $A,B$ are open sets forming a disconnection of   $\mathbb{I}$. Thus $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$ are   non-empty bounded disjoint sets whose union is $\mathbb{I}$. Since $A$   and $B$ are open, the sets $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$   cannot consist of only one point. (Why?) For the sake of definiteness,   we suppose that there exist points $a\in A$, $b\in B$ such that   $0&lt;a&lt;b&lt;1$. Applying the supremum property, we let $c=\sup\{x\in  A:x&lt;b\}$ so that $0&lt;c&lt;1$; hence $c\in A\cup B$. If $c\in A$ then $c\ne  b$ and since $A$ is open there is a point $a_1\in A$, $c&lt;a_1$, such   that the interval $[c,a_1]$ is contained in $\{x\in A: x&lt;b\}$ contrary   to the definition of $c$.   Why is $0&lt;c&lt;1$? They define $c=\sup\{x\in A:x&lt;b\}$ so it is the supremum of $A$ which is less than $b$ so doesn't that mean $c$ must be less than $b$ ? Also, why is $c\in A\cup B$ ? Lastly, since $c$ is the supremum of $A$ how is there a point in $A$ such that $c&lt;a_1$ and why is it contrary that it is contained in $\{x\in A: x&lt;b\}$ ?
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The definition of $T_1$-Space is:  A topological space $X$ is said to be $T_1$ if for each pair of distinct points $a,b, $ $\exists$ open sets $U,V$ s.t $a\in U, b\notin U, a\notin V, b\in V$.  What I'm confused about is in a $T_1$ space, all singleton subsets of $X$ are closed. Let $t,v \in X$. Then I think the singleton sets $\{t\}$ ,$\{v\}$ satisfy the definition of $T_1$ in $U$ and $V$ what I wrote above. (i.e $t \in\{t\}$, $v\notin \{t\}$, $t\notin\{v\}$, $v \in\{v\}$.) I learned the theorem showing this result and I can understand the proof of it, but I'm still confused as to why this is not a counterexample.
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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  $U^o=U$
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The definition of a boundary of a set $S$ in a topological space $X$ is $\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\}$ (complement of the interior union exterior).  The definition for interior is the set of all interior points.  The definition for interior point of $S$ is if there exists an open neighborhood $N$ of that point such that $N \subset S$.  The definition of neighborhood of a point $x$ is a subset of the topological space $X$ that contains an open set such that $x$ is in that open set.  The definition for exterior of $S$ is $\text{Int}(\text{Comp}(S))$.  So $\text{Bdry}(\text{Bdry}(S)) = \text{comp}\{\text{Int}(\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\}) \cup \text{Ext}(\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\})\}$ which I can't make heads or tails of. Is there an easier approach to checking of the boundary of the boundary is a subset of the boundary using these definitions?
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Prove that if a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected.   My professor once mentioned a proposition likewise but I cannot find the exact version of it. Can anyone help?
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Let $X$ be a banach space, and let $A$, and $B$ be closed linear subspaces. Assume that $$\inf\{\|x-y\|\mid x\in A, y\in B, \|x\|=\|y\|=1\}&gt;0$$  I want to show that $A+B$ is closed.   I was thinking of doing something like, let $z$ be a limit point of $A+B$, then there exist $z_n\in A+B$ such that $z_n\rightarrow z$, each $z_n$ can be written as $a_n+b_n$. Then I wanted to do something along the lines of determining whether $a_n$ and $b_n$ have a limit (if they do, then call them $a$ and $b$ and then $a\in A$, $b\in B$, and $z=a+b\in A+B$), but I can't seem to be able to use the condition given.   I am studying for a qual, so you can go ahead and either give a solution or sketch it.
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The following is problem 3.22 from Rudin's Princples of Mathematical Analysis:     Suppose $X$ is a nonempty complete metric space, and $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove Baire's theorem, namely, that $\bigcap_{n=1}^\infty G_n$ is not empty. Hint: find a shrinking sequence of neighbourhoods $E_n$ such that $\overline{E}_n\subset G_n$.   Here's what I've tried so far:      Let $\{r_n\}$ be a Cauchy sequence of positive real numbers converging to $0$. Fix $x\in X$ and define $E_i=\{g\in G_i:d(g,x)&lt;r_i\}$, which is nonempty since $G_i$ is dense. I would like to show that for all $i$, $\overline{E}_i\subset G_i$ (I had convinced myself that this would be true but I am now having doubts). Let $e\in \overline{E}_i$. Then either $e\in E_i$ or $e$ is a limit point of $E_i$. If $e\in E_i$ then $e\in G_i$. Otherwise, every neighbourhood of $e$ contains a point in $E_i$.    I thought that I should be able to then choose some point $e'\in E_i$ in a neighbourhood of $e$ and, since $G_i$ is open, it'll have a neighbourhood $N\subset G_i$ which contains $e$, but this is proving to be difficult and I'm worried that it's not true. If I can show that this is true then the rest will follow from results I've already proven.  Does my approach make any sense?  Incidentally, as a secondary question, what type of a thing would $G_n$ be? A sequence of dense open subsets seems weird to me—at first I was thinking of some sequence of infinite subsets of rational numbers in the real numbers but I realized that those aren't open. Is there anything which would be familiar to my little undergrad brain which would be analogous to this problem?
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My attempt to prove the following is below:     Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$   My book calls this set the ""  I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly)  Using the definition of topology:     $\mathscr{T}$ is a topology on X if and only if the following are true:   (i) X and Ø are elements of $\mathscr{T}$.   (ii) $\mathscr{T}$ is closed under finite intersections.   (iii) $\mathscr{T}$ is closed under arbitrary unions.   Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2...$ are all finite sets)  So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.   Now to verify the definition.     (i) X and Ø are elements of $\mathscr{T}$.   Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.      (ii) $\mathscr{T}$ is closed under finite intersections.   $U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| &lt; |\mathbb{N}|  $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.       (iii) $\mathscr{T}$ is closed under arbitrary unions.   $U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| &lt; |\mathbb{N}|  $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.  Now let's consider the infinite case: $V_k \cap ... \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap ... $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.
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I am confused about definition of the boundary set. Definition: Given set $A$ and its complement $A^c$; Point $x$ belongs to the boundary of $A$ if every open ball centred at $x$ contains points of $A$ and $A^c$.   Now to my confusion: Taking as an example set $A\subset \mathbb{N}$ and its complement in $\mathbb{R}^1$. Set $A$ does not contain any limit points. Then for $\forall x\in A$ the only member of $A$ that is contained in all the balls centered at $x$, is $x$ itself. Now, $x\notin A^c$ but is a limit point in $A^c$, hence every ball centered at $x$ contains elements of $A^c$. Can I conclude that $x$ is in the boundary set of $A$, even though it is an isolated point in $A$? Also, say if the complement of $A$ is also in $\mathbb{N}$, will $x$ still be a boundary point of $A$?  Thanks a lot in advance! And I apologize for the clumsy language :-)  Leon
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I have to prove that if $E$ is a connected space, then so is $\overline{E}$ (the closure of $E$) a connected space. I tried to prove the contrapositive. So suppose that $\overline{E}$ is not connected in a metric space $X$. This implies there are $A\subset X$ and $B\subset X$ such that $A\cap B=\emptyset$, $A\cap \overline{E}\neq \emptyset$, $B\cap \overline{E}\neq \emptyset$ and $\overline{E}\subset A\cup B$. I will try to prove that this $A$ and $B$ will also work for $E$. Suppose that $A\cap E=\emptyset$ and $B\cap E=\emptyset$, this implies that $A\cup B$ is a subset of the limit points of $E$, because $A\cap \overline{E}\neq \emptyset$ and $B\cap \overline{E}\neq \emptyset$. This contradicts the fact that $\overline{E}\subset A\cup B$, so we have to conclude that $A\cap E\neq \emptyset\neq B\cap E$. Since $E\subset\overline{E}\subset A\cup B$, we have that $E$ is also not connected.  My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance!   Edit1: The definition we use for a connected space $E$ in a metric space $X$ is that if $E$ is connected, then there are no open $A,B\subset X$ such that $A\cap B=\emptyset$, $A\cap E\neq \emptyset\neq B\cap E$ and $E\subset A\cup B$.  Edit2: $A$ and $B$ must be open.
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Let $X$ be a Hausdorff compact space; let $\{A_n\}$ be a countable collection of closed set of $X$. Show that if each set $A_n$ has empty interior in $X$, then the union $\bigcup A_n$ has empty interior in $X$.   My solution:  Suppose that, $U=$ interior of $\bigcup A_n$ is nonempty. Then any open subset of $U$ can not have isolated point, since then all interior of $A_n$ can not be empty. (If $U$ have isolated point, then for some $x\in X$ with $\{x\}\subset U$ is an open set, then it must be a subset of some $A_n$, since $\{x\}$ do not intersects with $X-A_n$).  Let, $x_1\in A_1$. Choose a point $x_2\in U$ that do not contain in $A_1$. Now choose disjoint sets $W_1$ and $W_2$ about $x_1$ and $x_2$. Now, take $V_1=W_2\cap U\cap A^c_1$. Then, $x_1$ do not contain in $\bar V_1$.  Again do this: There exist an $k_1$ such that $x_2\in A_{k_1}$. Choose a point $x_3\in V_1$ that do not contain in $A_{k_1}$. Now choose disjoint sets $W_3$ and $W_4$ about $x_2$ and $x_3$. Now, take $V_2=W_4\cap U\cap A^c_{k_1}$. Then, $x_2$ do not contain in $\bar V_2$.  In this way we can find a collection of closed sets $\{\bar V_n\}$ such that $$\bar V_1\supset \bar V_2\supset\bar V_3\supset\dots$$  Note that, since $X$ is compact, $\bigcap \bar V_n$ is nonempty. Let, $x\in \bigcap \bar V_n$.   Interior of $A_n$ is empty and $x\notin A_n$ for all $n$ leads to a contradiction.  Is this solution ok? Any help appreciated.  Sorry that I published same question twice. I marked the  duplicate.