I am self studying Munkres' Topology and was struck on this exercise question of section 13. I don't have any help from teacher or friends.  So, I am asking for any explanation here.  The question is     Show that if $A$ is a basis for a topology on $X$, then the topology generated by $A$ equals the intersection of all topologies on $X$ that contain $A$.   Prove the same if $A$ is a subbasis.   The intuition is clear to me but I don't know how to write a proper proof.  Can you please tell me for both parts.
121900
Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$.   This is what I've managed. However, the following stumps me:   Prove the same if $\mathscr{A}$ is a subbasis.   That is, how to show that the topology generated by $\mathscr{A}$ as a subbasis is the same as the intersection of all topologies on $X$ that contain $\mathscr{A}$?   Let $\mathscr{T}$ be the topology that consists of all unions of finite intersections of sets in $\mathscr{A}$, and let $\mathscr{T}^\prime$ be the topology that is the intersection of all topologies that contain $\mathscr{A}$.   Now $\mathscr{A}$ is a collection of subsets of $X$ whose union is all of $X$, and each set $A \in \mathscr{A}$ is in both $\mathscr{T}$ and $\mathscr{T}^\prime$.   Am I right so far?   And if so, then what next?
69023
I am having a hard time understanding why the box topology is finer the the product topology. Of course I know that with finite product, the two are the same. With product topology, the basis elements are defined so that the $U_{\alpha}=X_\alpha$ except for finitely many $\alpha$. So how can this topology be contained in the box topology which has only the product of the  $U_{\alpha}$ as basis. I think it should have been the other way round. Am I missing something here? I need help with this because I have been confused over it for a long time.
99142
Let $X$ and $Y$ be two topological spaces.   Prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.   I am stuck on the converse. Suppose $f: X \to Y$ is continuous. Then for every closed set C in Y, $f^{-1}(C)$ is closed in X.   WTS  for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.     Since $\overline{f(A)}$ is a closed set in Y, $f^{-1} \circ \overline{f(A)}$ = $ \overline{ f^{-1} \circ \overline{f(A)} }$  Also, $f(\overline{A}) \subseteq \overline{ f(\overline{A})}$
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Could you please verify this (rather simple) proof? I'm a bit new to this kind of reasoning.  The question is:  Let $\mathcal{T}_1$, $\mathcal{T}_2$ be two topologies on some set $X$, when is the identity map $id:X \to X$, $id(x) = x$ continuous?  Here is my reasoning:   If $\mathcal{T}_1 = \mathcal{T}_2$, then for any $S \in \mathcal{T}_2$, $id^{-1} (S) = S \in \mathcal{T}_1$ so then it's continuous.  If $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it's exactly like above, then it's also continouos.  Now:  If  $\mathcal{T}_1 \subseteq \mathcal{T}_2$ then take some $x \in U \in \mathcal{T}_2 \setminus \mathcal{T}_1$, then $id^{-1}(U) \not\in \mathcal{T}_1$, so it's not continuous.  So the identity map is continuous if $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.   Is this OK?  Is there a better way to show this?  Any extra applicable theory?  Thanks in advance
65126
Prove or find counterexamples.  Let $X$ be an infinite set and $T$ be a topology on $X$. If $T$ contains every infinite subset of $X$, then $T$ is the discrete topology.
103693
I'm reading Intro to Topology from Mendelson.  The entire problem statement is,  Prove that $X$ is compact if and only if for each family $\lbrace F_\alpha\rbrace_{\alpha\in I}$ of closed subsets of $X$ that has the finite intersection property, we have $\bigcap_{\alpha\in I} F_\alpha\neq\varnothing$.  My attempt at the proof is as follows:  First assume that $X$ is compact. For the sake of contradiction, suppose that there exists a family $\{ F_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ with FIP such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$. Since $X$ is compact if $\lbrace F_\alpha\rbrace_{\alpha\in I}$ any family of closed sets such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$, then there exists a finite set of indices $\lbrace \alpha_1,\dots,\alpha_n\rbrace$ such that $\bigcap\limits_{i=1}^n F_{\alpha_i}=\varnothing$. Yet, since $\lbrace F_\alpha\rbrace_{\alpha\in I}$ has the finite intersection property, for every finite index of $J\subset I$, $\bigcap\limits_{\alpha\in I} F_\alpha\neq\varnothing$.  Suppose now that $X$ is not compact, that is, there exists an open cover $\lbrace U_\alpha\rbrace_{\alpha\in I}$ of $X$ with no finite subcover. Consider the set $F_\alpha=C(U_\alpha).$ Then $\lbrace F_\alpha\rbrace_{\alpha\in I}$ is a collection of closed subsets of $X$ with the finite intersection property and were $\bigcap_{\alpha\in I} F_\alpha=\varnothing.$  Thanks for any feedback!
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Consider Theorem 19.1 from Munkres' topology:     The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.   I am having a hard time seeing how the last sentence follows from the subbasis-definition of the product topology. I don't find Munkres' preceding explanation to be helpful.
16420
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.
109732
A topological space is locally compact then there is an open base at each point has all of its set with compact closure, how can i prove it?  It seems very meaningful when the topological space is metrizable and the topology is the topology induced by the metric, but i could not prove it for general topological spaces.  I am not familiarized with Hausdorff spaces.  What i tried for now is starting from the definition of locally compact as the topological space whose ever point has a neighbourhood with compact closure. I picked (for each point) this neighbourhood, and created a class with all open sets contained in that neighbourhood, then i thought that this class may be the open base at the analized point whose all sets have compact closure, but i could not prove that this class is an open base at point, have someone an idea? Or other path to prove it, or yet is this claiming false?
99412
Consider a topological group $G$. I want to prove that      if $\cal{U}$ is the set of all neighborhoods of $e$ (the identity element), then $\overline{\{e\}}=\cap_{U\in \cal{U}}U$.   What I thought was:  i) If $x\in \overline{\{e\}}$, then every neighborhood $V$ of $x$ intersects $\{e\}$, hence every $V$ is a neighborhood of $e$ as well, i.e. $V\in \cal{U}$.  ii) On the other hand, if $x\in \cap_{U\in \cal{U}}U$, then $x\in U$ for all $U \in \cal{U}$, meaning every $U$ is a neighborhood of $x$ intersecting $\{e\}$.  Now, I think that the missing part is an argument justifying that there is no neighborhood of $e$ that is not a neighborhood of $x$ (part i)), and that there is no neighborhood of $x$ that is not a neighborhood of $e$ (part ii)). My idea for this is to argue that there is a homeomorphism taking $e$ to $x$, therefore the neighborhood system of one cannot have more elements that the neighborhood system of the other.  Is this correct? If not, is there a way I can fix it?
16363
Hello I am doing this exercise where we let X and Y be topological spaces and $f:X\to Y$ a function. We suppose that $\mathcal{A}$ is a collection of closed subsets of X, such that $\bigcup_{A\in\mathcal{A}} A = X$. We assume that for every $A\in \mathcal{A}$ the restriction $f_{|A}: A\to Y$ of f to A is continuous. Furthermore we suppose that for each $x\in X$ there exists a neighborhood U of x such that the intersection $A\cap U$ is non-empty for only finitely many $A\in\mathcal{A}$. And I have to show that f is continuous. I know that since $f_{|A}: A\to Y$ is continuous we have that for all open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of A. And to show that $f: X\to Y$ is continuous I am thinking of showing that for each open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of X, but this would be obvious wouldn't it since for all open $O\subseteq Y$ we have $f^{-1}(O)\subseteq A\subseteq X$?
103829
PROBLEM Let $A$ be a dense subset of $X$ and suppose $A$ is connected in the induced subspace topology, then $X$ is also connected  ATTEMPT Suppose $X = U \cup V $ for open, nonempty, disjoint $U,V$. By Denseness of $A$, $Cl(A) = U \cup V $. We know that $A \subseteq Cl(A)$. Hence $A \subseteq U \cup V $. Well, here it is pretty obvious that $A $ must be $U \cup V$ and hence having a contradiction. But I think probably I need to prove this obvious thing. Can someone help me? Is this approach correct? Thanks,
77137
I'm doing exercise from "Topology without tears book", page 27, exercises a and b.   First one (a) reads as follows. "Let $\mathbb{R}$ be the set of all real numbers. Prove that each of the following collections of subsets of $\mathbb{R}$ is a topology.  a. $\tau_1$ consists of $\mathbb{R}$ , $\emptyset$ and every interval $(-n;n)$ for $n$ any positive integer  b. $\tau_1$ consists of $\mathbb{R}$ , $\emptyset$ and every interval $[-n;n]$ for $n$ any positive integer"  To prove this I noticied that     $S_n=(-n;n) \in \tau_1 \; \forall n \in \mathbb{N}$ by definition.   $S_{n_1} \subset S_{n_2} \; \forall \; n_1 &lt; n_2, n_1,\;n_2 \in \mathbb{N}$   From 2 it follows that    $S_{n_1} \cap S_{n_2} = S_{n_1} \; \forall n_1 &lt; n_2 $  $\bigcup_{i=1}^{i=n} S_i = S_n \in S_{n+1} $   Sience intersection of any two sets is in $\tau_1$ and union of any number of sets is also in $\tau_1$ and noticing that $S_n \subset \mathbb{R} \; \forall n \in \mathbb{N}$ I can state that $\tau_1$ is a topology of $\mathbb{R}$  Is this proven right? I have the following question. Exercise b is very similar to exercise a - the only difference is that now $[-n; n]$ is a closed interval. I think that I still can use the same prove for this case too, but not sure - does closed interval $[-n;n]$ changes anything?
16385
I'm reading topology of Munkres and I have a problem that stuck me for a while. I'm so greatful if anyone can help me with this.     Let $A$ be a proper subset of $X$, and let $B$ is a proper subset of $Y$. If $X$ and $Y$ are connected, show that $$(X\times Y) \setminus (A\times B)$$ is connected.   Thanks so much for your consideration ^^
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.
10564
You have two connected topological spaces $(A,B)$. Prove that $A\times B$ is also connected.  I understand that I have to prove that there is a point in $B$ (call it $b$), that makes $A\times\{b\}$ homeomorphic to $A$ making it connected to $A\times B$. Then prove that $\{a\}\times B$ is connected in $A\times B$. But I don't really know where to being with this. If you could help that would be appreciated.
10116
I'm being asked to prove that $\mathcal{A}=\left\lbrace A_{a,b}:a,b\in\mathbb{Z}\right\rbrace$ is a basis for a topology on $\mathbb{Z}$, being: $$A_{a,b}=\left\lbrace a+nb:n\in\mathbb{Z}\right\rbrace=\left\lbrace ...,a-2b,a-b,a,a+b,a+2b,...\right\rbrace$$  Obviously every $a\in \mathbb{Z}$ belongs to a basis element, for example $A_{a,b}$, for any $b$. I'm having difficulties though, proving that if two of those sets $A_{a,b}$ and $A_{a',b'}$ share an element $x$, then there exists another $A_{c,d}$, such that $x\in A_{c,d}$ and $A_{c,d}\subset A_{a,b},A_{a',b'}$. Any hint?