Let $X$ be a normal space. By definition it is a $T_1$ space and any pair of disjoint closed sets can be separated by open sets in the sense that they have disjoint neighborhoods. What I want to prove is  $X$ is normal iff each nbd of a closed set contains the closure of some nbd of the set  To prove the first implication I took a closed set $A$ and nbd $U$ of $A$. Since $U'$ is closed and disjoint from $A$ there exist nbd $H_A$ of $A$ and nbd $H_{U'}$ of $U'$ such that they are disjoint. I feel that closure of $H_A$ is subset of $U$ but not able to prove it. For converse part, let $A$ and $B$ be closed disjoint sets in $X$. Then for any nbd $U_A$ and $U_B$ of $A$ and $B$ respectively there exist nbd $V_A$ and $V_B$ of $A$ and $B$ such that $$ A \subseteq V_A \subseteq \bar{V_A} \subseteq U_A $$ and $$ B \subseteq V_B \subseteq \bar{V_B} \subseteq U_B $$ If $U_A$ and $U_A$ are disjoint then we are done. If not how can I proceed further? Please suggest any alternative approach also if possible.
121007
Let $X$ be a $T_1$-space, and show that $X$ is normal if and only if each neighbourhood of a closed set $F$ contains the closure of some neighbourhood of $F$.  I saw this statement in the book Introduction to topology and modern analysis by Simmons. I tried to prove it but I did not get any where. How may it be proved?
36024
Suppose a $T_4$ space $X$ satisfies a property that every continuous map $f$ from $X$ to any Hausdorff space $Y$ is a closed map. Show that $X$ is compact. What I am thinking is to let $\mathcal{U}$ be an open cover of $X$ and find a way to choose finite open sets from $\mathcal{U}$ that can cover $X$ again. Do we need to define a suitable continuous map by ourselves and use the fact that it is a closed map to choose suitable open sets from $\mathcal{U}$? I have tried to define a continuous map $f \colon X \to Y$ as mapping different open sets from $\mathcal{U}$ to different points in $Y$. However, different open sets from $\mathcal{U}$ may intersect with each other and hence $f$ may not be well-defined. Apart form that, I have no idea about how to choose suitable $f$ and $Y$. Can someone help me please? Edit 1: I am trying to apply the Urysohn Lemma or the Tietz Extension Theorem due to the normality of $X$. But still have no idea. Edit 2: Finally found a way to prove without using compactification. I will post the proof when I’m free. Hint: Use Urysohn Lemma and Tychonoff Theorem.
92686
Fact  The intersection of a closed set $A$ and a compact set $B$ is closed.  Background  I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.  Relevant Facts  I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.  I also don't have that $B$ is closed otherwise it would be trivial.  Some Clarification  I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer      So if $K$ is compact in an arbitrary topological space (which is just to say that   it is a compact topological space when given its induced topology) and     $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology   Is it true if $A$ and $B$ are not whole spaces?  I would appreciate any hints.
121049
Show that for any topological space $X$ the following are equivalent:         $X$ is extremally disconnected   Every two disjoint open sets in $X$ have disjoint closures.      My attempt at a solution  Assume every two disjoint open sets in $X$ have disjoint closures. Let A open in $X$, $A$ and $X \setminus \overline{A} $  disjoint open set $\overline{A} \cap (\overline{X \setminus \overline{A}} )= \emptyset$,  $(\overline{X \setminus \overline{A}} ) \subset X \setminus \overline{A}$, which implies that $X \setminus \overline{A}$ is closed in $X$. Thus $\overline{A}$ is open in $X$. So we obtain $X$ is extremally disconnected.  Otherwise Let $X$ is extremally disconnected. Then the closure of every open set is open.  So $X$ is completely separated. Let $A$, $B$ two disjoint open sets.  Then there exist a continuous function $g:X \rightarrow [0,1]$ such that  $g(A)=0$, $g(B)=1$  How can I continue?
41884
If $\{E_\alpha\}_{\alpha\in A}$ is connected in $\mathbb{R}^n$, $\bigcap\limits_{\alpha\in A}E_\alpha \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E_\alpha$ is connected.  I have zero intuition on how to do a proof of this statement. Please help.  I started it by contradiction like this: suppose that $\bigcup\limits_{\alpha\in A}E_\alpha$ is not connected. Then there exist non-empty disjoint separating sets $U$ and $V$, such that $U$ and $V$ are relatively open in $\bigcup\limits_{\alpha\in A}E_\alpha$, and $U\cup V= \bigcup\limits_{\alpha\in A}E_\alpha$. But I have no idea what to do next.
91328
$p : X \to Y$ is continuous, closed and surjective, and $X$ is a normal space. Show $Y$ is normal.   There is a hint, which I'm trying to prove: show that if $U$ is open in $X$ and $p^{-1}(\{y\}) \subset U$, $y \in Y$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W) \subset U$.  I have a candidate for $W$, namely $W=Y\setminus p(X \setminus U)$. I did prove that this $W$ is open, and that $p^{-1}(W) \subset U$, but I don't see how $y \in W$. I think this would require injectivity of $p$.  I have also shown that $y \in p(U)$ and that $W \subset p(U)$, so if also $W \supset p(U)$, then $y \in W$.   Can anyone help me?
27350
Let $X$ denote a topological space. For any $A \subseteq X$, consider two possible conditions on $A$.   $A$ is closed $A \cap K$ is closed, for all compact $K \subseteq X$.   If $X$ is Hausdorff, then clearly (1) implies (2). If $X$ is metrizable, then the converse holds; see .     Questions.         What are some Hausdorff topological spaces in which (2) does not   imply (1)?   For Hausdorff topological spaces: what weaker conditions than metrizability are known to imply that (2) implies (1)? Are there any interesting characterizations of those Hausdorff spaces in which (2) implies (1)?
28137
I would like to prove the following statement.  Let A and B be nonempty disjoint subsets of $\mathbb{R}$ where A is closed and B is compact. Then there exists $a \in A$ and $b \in B$ such that $inf${$|a-b|$:$a \in A, b \in B$}=$|a-b|$    I read a similar question above where the concept of "limit point" and others were used, but my class (Introduction to Analysis) has not gone that far; most of all, I would like to prove the statement from the knowledge that the class covered.  So far, I have learned definition of limit of a sequence, Bolzano-Weierstrass Theorem. Also, I just learned (and proved) that there exists a sequence ${a_n}$ in E such that $\lim{a_n} = \inf{E}$ if E is a nonempty subset of $\mathbb{R}$ (whether E is bounded below or not due to the extended real numbers).   From the things that I learned, first, I constructed a set $E$={$|a-b|$:$a\in{A}, b\in{B}$} and I know that there exists a sequence ${x_{n}}$ in E such that $\lim{x_{n}}=infE$. Since ${x_{n}} \in{E}, x_{n}=a_{n}-b_{n}$ where $a_{n} \in A, b_{n} \in B$. Let $infE = \beta$. Then, this leads to $\lim{(a_{n}-b_{n})}=\beta$. Then, for any $\epsilon$>0, there exists $N \in$ $\mathbb{N}$ such that $|a_{n}-b_{n}-\beta|&lt;\epsilon$ for all n $\ge$ N.   Then, I could not proceed. I should use conditions that A and B are disjoint, A is closed, and B is compact. Could anyone help me proceed? I had a feeling that using contradiction might work from the link above, but it is quite tough. Since I am a novice (just started taking Introduction to Analysis), I would very much appreciate it if you could help me prove this at the beginner level.
119394
Suppose that $V, W$ are closed in $X$ and that $V\cap W, V\cup W$ are connected. Show that $V$ and $W$ are connected.  My attempt: Suppose that $V=U_1\cup U_2$ where $U_i$ are open in $V$, not empty and $U_1\cap U_2=\emptyset$. This induces a separation for $V\cap W$: $$ V\cap W = (U_1\cap W)\cup(U_2\cap W).$$  $U_i\cap W$ is open in $V\cap W$  $(U_1\cap W)\cup (U_2\cap W)=\emptyset$ as otherwise $U_1\cap U_2 \ne \emptyset$.  Now, I need to show that both sets are not empty. We know that $V\cup W$ is connected and closed in $X$. So, for all $x\in V\cup W=U_1\cup U_2\cup W$ and all neighborhoods $Z$ (in $X$!) of $x$ we have that $Z\cap (V\cup W) \ne \emptyset$. We can choose $x\in U_1 \subseteq V\cup W$. But then I would have to find a neighborhood of $x$ in $X$. I can't find one... How can I complete this bullet point?   Thanks.
12018
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.    This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
76505
Let $X$ be a topological space. Let $A$ and $B$ be two subsets of $X$ such that: $A$ and $B$ are closed, $A \cup B$ is connected, $A \cap B$ is connected     Prove that $A$ and $B$ are connected.   Thoughts:  Suppose that $A$ is disconnected. Since $A$ is closed, there are two non-empty, disjoint, closed sets $E$ and $F$ in $X$ such that $A = E \cup F$. Then I attempted many things but all were not really useful.  Please help. Thanks.
46729
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?
4433
Let $X$ be a topological space.  It is  that, if $X$ is compact, then every proper closed subset $Y \subset X$ is compact.  Out of curiosity, I would like to explore the converse of this statement:     Let $X$ be a topological space with the property that every proper closed subset is compact.  Is $X$ itself compact?   Of course, this question is only interesting when $X$ is infinite.  We can further restrict ourselves to spaces that have infinitely many open sets since every closed set, and indeed the space itself, would be compact otherwise.  If this result is false in general, what hypotheses can be added to make it so?  Should this not hold, I'm interested in finding concrete examples of when it does.  For instance, we can take any infinite set endowed with the , which works because every closed set is finite and thus compact, and the space itself would be compact.  Another example is an interval $[a, b] \subset \mathbb{R}$, wherein closed subsets of this are closed and bounded in $\mathbb{R}$; as such, they are compact per .    Update:  Zardo's answer reveals that such a topological space is indeed compact. For further discussion on the  and its relevance to compactness, see Munkres' : Chapter $3$, theorem $26.9$.
25264
Let $X$ be topological space and $A\subseteq X$.   If $E\subseteq X$ is connected, $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$.  Assume that $E\cap\partial A=\emptyset$.  If $E\cap\partial A=\varnothing$ then $E=(E\cap\text{int}(A))\cup(E\cap\text{ext}(A))$, i.e. a union of disjoint sets open in $E$.   Also the sets are nonempty because: $E\cap\text{int}(A)=E\cap A\neq\varnothing$ and $E\cap\text{ext}(A)=E\cap A^c\neq\varnothing$.  This contradicts that $E$ is connected.  Fixed.
73178
I'm trying to prove $(A \cup B)' = A' \cup B'$ where $S'$ denotes the derived set of some subset $S$ of a topological space $X$.  A derived set $S'$ of a set $S$ is the set of $x \in X$ such that $x$ is in the closure of $S-\{x\}$.  I'm having troubling showing that $(A \cup B)' \subseteq A' \cup B'$.   What I've done so far: Suppose $x \in (A \cup B)'$.  Then for any neighborhood $U$ of $x$ it follows that $U$ intersects $(A \cup B)-\{x\}$ so $U$ intersects $A-\{x\}$ or $B-\{x\}$.  But I don't see why we couldn't have one neighborhood of $x$, $U_{1}$, which intersects $A-\{x\}$ but not $B-\{x\}$ and another neighborhood $U_{2}$ which does the opposite.  In this case $x$ would be in neither $A'$ nor $B'$.  We can say that $U_{1} \cap U_{2}$ is not only $\{x\}$ because then $\{x\}$ would be open contradicting $x \in (A \cup B)'$.
88050
A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.   Prove that every separable metric space has a countable base.   Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x&gt;0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)&lt;{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.   It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)&lt;\varepsilon_x/4$ and since $d(x, d_j)&lt;{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)&lt;{\varepsilon_x}/{4}+{\varepsilon_x}/{4}&lt;{\varepsilon_x}.$$  Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.  By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r&lt;{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.  Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls  arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.  Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.  Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.
65819
I was studying a text () when in a proof the author stated:  !  If I got it right, the author says that $\widetilde{U}$ (a nonempty open connected set in $\mathbb{C}^n$) is a finite union of closed sets, so one of these closed sets must have nonempty interior.   I tried to prove it but I could not. I think it is a very simple topological detail I am missing.