I have to prove this:  Let $(X,T)$ be a topological space , then it is equivalent:  $1)$  The closure of an open set is an open set.  $2)$  If $U$ and $V$ are disjoint open sets, their closures are also disjoint.  I proved $ 1\Rightarrow 2$, but I see no way of doing the other one. Any help would be helful with this.
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?
23367
As a part of self study, I am trying to prove the following statement:  Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.   Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
94054
I have seen two definitions of locally compact topological space   A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists a compact set that contains an open neighborhood of $x$. A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists an open neighborhood of $x$ whose closure is compact.   Are these two definitions equivalent? It's clear that the second implies the first, but I don't see why the first implies the second. Suppose that $U $ is an open neighborhood of $x$ with $x\in U\subset K$, here $K$ is a compact set. Since compact sets may not be closed in a non-Hasudorff topological space, the closure of $U$ may not be contained $K$, so we can't say the closure of $U$ is compact.
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I'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.  "Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."  Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.  Thanks.
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Claim: If $(X,\tau)$ is not $T_1$ then $A'$ is not necessarily closed for any $A$.  Proof: Here is an example. Take $X=\mathbb{N}$ and let $\tau=\lbrace u\subset\mathbb{N}\mid 1\in u\iff 2\in u\rbrace$. Let $A=\lbrace 2,3 \rbrace$. Since any open set containing $1$ also contains $2$ it follows that $1\in A'$. Hence $\overline{A}=\lbrace1,2,3\rbrace$, as it is the smallest closed set containing $A$. But  $X\setminus A'=\mathbb{N}\setminus\lbrace 1\rbrace$ is not open because it contains $2$ and not $1$. The set of accumulation points $A'$ is not closed.  So, in general, trying to prove that $A'$ is closed in a topological space is an effort in futility?
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Give an example of a topological space $(X, \mathcal T)$, a subspace $(A, \mathcal T_A)$ of $(X, \mathcal T)$, and a closed set in $(A, \mathcal T_A)$ that is not closed in $(X, \mathcal T)$.   All the sets I create are closed in both and I am stuck.
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I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.
97688
Hi I just want to find out if the following is a acceptable proof for the proposition:   "Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."  Proof:  Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$,  $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.   W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.   $\square$
111463
Theorem A map $\phi$ maps a topological space $X$ to another $Y$, where $X$ is Hausdorff, $Y$ is compact and the graph of $\phi$ is closed. Then $\phi$ is continuous.   Is it reallly necessary to include the condition that $X$ is Hausdorff?  Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer.            A proof:     For any closed subspace $C$ of $Y$, the pre-image $D$ ought to be closed. For any element $a$ in the complement of $D$, we can use the compactness to show that there is a finite number of open sets in $Y$ such that the union of them covers the C, and then the corresponding open sets in $X$ is an open neighborhood of $a$ which has an empty intersection with $D$; so $D$ is closed.     P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that $(a,c)$ is not in the graph for any $c$ in $C$.
94530
Show that every open set A in a metric space (X,d) is the union of closed sets.  I am beyond confused for this question. I thought union of closed sets are closed
3875
I'm really confused with the statement.   let $X$ be a set endowed with the discrete topology and let $Y$ be any topological space. Show that $M(X,Y)$, the space of continuous maps from $X$ to $Y$ endowed with the compact open topology, is homeomorphic to the topological product $ \prod_{x \in X} Y_x$, $Y_x = Y$.  Can you please explain what we really have to prove?
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It is a Dugundji's exercise:  Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.  If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.  For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$.   Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.
69969
Let $(X,\tau)$ be a topological space, $U \subseteq X$ a non-empty open subset of $X$ and let $W$ be an irreducible non-empty closed subset of $X$. Assuming $W \cap U \neq \emptyset$ why is this intersection irreducible in $U$ and its closure (taken in $X$) equal $W$?  Clearly $W \cap U$ is dense in $W$ because it is an open subset of $W$ and $W$ is irreducible so its closure, taken in $W$ is $W$ itself. I don't see why its closure taken in $X$ is equal $W$. Why is this? Also, why is it irreducible in $U$?  EDIT: OK $W=cl_{X}(W)=cl_{X}(W \cap U)$ I think. Why is it irreducible in $U$ though?
93637
Let $X$ be a topological space and $A\subseteq X$.  I have to prove that if $der(\{x\})$ is closed and $x\in der(der(A))$ then $x\in der(A)$.  I tried this... if $x\notin der(A)$, there exists an open set $U$ with $x\in U$ such that $U\cap A=\emptyset$. We know that $x\in X\setminus der (\{x\})$ and this is open, then $x\in U\cap (X\setminus der\{x\})$ and this is open. Maybe this will imply a contradiction... What do you think?  Edit: We recall that $der(A)$ denotes the set of accumulation points of $A$ (some people say "limit point" instead "accumulation point")  Thanks.
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Let be X a topological space and let be $Z,Y\subseteq X$ such that $Z\subseteq Y$: so if Z is compact in Y then Z is compact in X.  To prove the assertion let be U an open cover of Z in X and we consider the open cover $\mathcal{U}'=\{U\cap Y:U\in\mathcal{U}\}$ of $Z$ in $Y$ -indeed $Z\subseteq\bigcup\mathcal{U}=\bigcup_{U\in\mathcal{U}}U\rightarrow Z=Z\cap Y\subseteq(\bigcup_{U\in\mathcal{U}}U)\cap Y=\bigcup_{U\in\mathcal{U}}(U\cap Y)=\bigcup\mathcal{U}'$- thus there exist $\mathcal{U}''\subseteq\mathcal{U}'$ such that $|\mathcal{U}''|&lt;ℵ_0$ and $Z\subseteq\bigcup\mathcal{U}''$ and so $\mathcal{U'''}=\{U\in\mathcal{U}:U\cap Y\in\mathcal{U}''\}$ is an open and finite cover of Z in X -indeed $|\mathcal{U}'''|=|\mathcal{U}''|&lt;\aleph_0$ and $Z\subseteq\bigcup\mathcal{U}''\subseteq\bigcup\mathcal{U}'''$- and so this prove that $Z$ is compact in $X$.  It is my proof correct? if not, how prove the assertion? could be that if is false?  Could someone help me, please?
55022
When a lattice $(L,\leqslant)$ is a lattice of open (or closed) sets of some topological space $(X,\tau)$? Which conditions have to be satisfied? We may assume that $X$ is $T_1$.
97034
I'm trying to do the following:     Prove that a function $f : S\rightarrow T$ between two topological spaces is continuous iff $f^{-1}(C)$ is closed whenever $C\subset T$ is closed.   To show that the closed-set relationship implies continuity, notice that $T-C$ is open. The preimage of this open set is $f^{-1}\left(T-C\right)=S-f^{-1}(C)$, which is open by assumption ($f^{-1}(C)$ is closed). Of course, any open set $G\subset T$ can be written in the form $T-C$ by letting $C=T-G$. Thus, for every open set $G\subset T$, the preimage is open.  To show the implication in the other way, consider a closed set $C\subset T$. $T-C$ is open, so $f^{-1}(T-C)=S-f^{-1}(C)$ is open as well ($f$ is continuous), which means that $f^{-1}(C)$ is closed.