Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space,  Hausdorff?     I think it is true since if $a $ and $b $ are two distinct points, they can be separated two disjoint  open sets, since the main space is totally disconnected (see Theorem at &lt;  >).   Is this argument true?
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The Wikipedia article on totally disconnected spaces seems to imply they are not necessarily Hausdorff (they are all $T_1$ though). What's an example of a totally disconnected non $T_2$ space?  (A space is totally disconnected if all connected components are singletons.)
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Question is :  Suppose $A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$ then    $A$ is compact Every subset of $A$ must be compact $A$ must be finite $A$ is disconnected   Only compact subsets of $\mathbb{R}$ I can think of are finite union of closed intervals and finite sets.  Take a  finite union of closed intervals, If i remove one element for that I would end up with something which contains $[a,b)$  and this would be not compact so should be the whole set.  So, First option is wrong i.e., $A$ is not compact.  I can not say anything about second and fourth option but third option is possible I guess.  Please help me to see this in detail..
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We know that, the intersection of arbitrary many compact sets is compact in Hausdorff Spaces.  I need counter example analogous of above proposition for the union is not true. When I consider, logically it does not have to be compact but I cannot find a counter example in one way or another. Could someone give a counter example please? Thanks
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It is just an idea (might be wrong) but, i think that if a Hausdorff space, say $X$, contains too many elements, then a countable subset cannot be dense in it.  Does there exist a cardinality that any space with that (or bigger) cardinality cannot have a countable dense subset?  Thanks.
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I am working on the problems from the textbook "Topology without tears". I am stuck with problem number $4$ in Exercise $2.2$. Could anyone suggest some hints on how to proceed? The question goes as follows.  A topological space $(X,\tau)$ is said to satisfy the second axiom of countability or to be second countable if there exists a basis $\mathcal{B}$ for $\tau$, where $\mathcal{B}$ consists of only a countable number of sets.   Prove that the discrete topology on an uncountable set does not satisfy the second axiom of countability. Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfy the second axiom of countability?   For the first problem, I tried to argue by contradiction assuming that the basis for the topology is countable. But I did-not know what I need to look at to prove the contradiction. For the second one, I do not know where to start. I have not thought deeply on the second one though.  Thanks, Adhvaitha
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If a metrizable topological space is totally bounded with one metric, is it totally bounded with all others? A related, stronger question: if every metrization of a topological space is bounded, are they all totally bounded?
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Let $(X,d)$ be a metric space. Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$ such that $E\cap A\neq\emptyset, E\cap B\neq \emptyset$ and $E\subset A\cup B$.   I'm not sure how to begin, so let me just start by pointing some stuff out from the question that I noticed, and my knowledge as of now. (Hopefully it'll be useful.)   There is $a\in A, b\in B$ which are both limit points of $E$ (I think) Since $A$ and $B$ are disjoint, $A\cap B=\emptyset$  The closure of a set is the "smallest" set which contains its limit points, (so it is a closed set as well).  Both $A$ and $B$ are open so they are made up entirely of interior points, so they don't have any points on the boundary.   I know that the definition of separated means that I have $A,B\subset X$, for which $A\cap \bar{B}=\emptyset=\bar{A}\cap B$, where the bar above the set denotes its closure.  And I know that the definition of connected set means for $E\subset X$, $E$ is connected if $E$ is not the union of two non-empty separated sets, so a disconnected set would be for $E\subset X$, $E$   is disconnected if it is the union of two separated sets (which are both non-empty).  Any hints as to how to begin would be appreciated. Thank you.
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Let $X$ be a normal Hausdorff space and let $C,D$ be two $F_{\sigma}$ subsets of $X$ such that $\overline{C} \cap D = \emptyset$ and $C \cap \overline{D} = \emptyset$.  Prove there exists disjoint open subsets $U,V$ such that $C \subset U$ and $D \subset V$.  No idea how to show this. Can you please help?
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Let $X$ be a metric space ; then which of the following is possible ?  1) $X$ has exactly $3$ dense subsets   2)  $X$ has exactly $4$ dense subsets   3)  $X$ has exactly $5$ dense subsets   4)  $X$ has exactly $6$ dense subsets   I know that if $X$ has a proper dense subset then for some $a \in X$ , we should have $X \setminus \{a\}$ is dense in $X$ and then $\{a\}$ is not open in $X$ ; but I can't relate this to no. of dense subsets except that if $X$ has only finitely many dense subsets then the topology of $X$ cannot be discrete . Please help . Thanks in advance
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Let $X$ be a Hausdorff, locally compact but non-compact topological space. If the (Alexandroff) one-point compactification is connected, can $X$ have compact connected components?
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The following are simple observations.  Suppose $\mathcal{T}_1,\mathcal{T}_2$ are two topologies on a set $X$ such that $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.  If $( X ,\mathcal{T}_2 )$ is Hausdorff, then so is $( X ,\mathcal{T}_1 )$. If $( X ,\mathcal{T}_1 )$ is connected,  then so is $( X ,\mathcal{T}_2 )$.   The real numbers with the usual topology is both Hausdorff and connected, which leads to the following two questions.  Is there a topology on the real numbers strictly coarser than the usual one, which makes it still Hausdorff?  Is there a topology on the real numbers strictly finer than the usual one, which makes it still a connected space?
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I've seen a few questions already posted on here, but they all deal with perfect subsets and being Hausdorff, both of which are topics we haven't covered yet.  I know that a point a is isolated if $\{a\}$ is open (aka there exists $\epsilon &gt; 0$ s.t. $B_{\epsilon}(a) = \{a\}$.  I am supposed to consider singletons and then use Baire Category Theorem.  Could someone help me with where to start?
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Question: Let $(X, d)$ be a metric space. Let $A$ and $B$ be disjoint subsets of $X$ that are closed in the topology induced by $d$. Prove that there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B\subset V$.   Is this proof correct? : By assumption, $A\cap B=\emptyset$, so there is no pair like $(x_A,x_B)$ (i.e., $x_A\in A$ and $x_B\in B$), such that $x_A=x_B$. Because every metric space is Hausdorff, so for every pair $(x_A,x_B)$ there are two disjoint open sets $U_A\ni x_A$ and $V_B\ni x_B$; and thus union of all $U_A$s (which is an open set containing A) are disjoint from union of all $V_B$s (which is an open set containing A).   Thanks a lot for any help.    PS In special case of standard metric topology on $\mathbb{R}$, if $[a,b]$ and $[c,d]$ are disjoint (say $c&gt;b$), intuitively it is apparent that there is a 'space' between $c$ and $b$ so we can have two disjoint open sets, each of them containing $[a,b]$ or $[c,d]$.
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Moderator's note: I've copied the title of the post from a more recently posted version of this question by the OP. The text below is from the original question: I am not sure whether the motivations still applied, hence I left it untouched.   It is well known that any connected metrizable space with more than two points is uncountable. And any connected metrizable space has infinitely many open spheres; as each such sphere is connected, there are infinitely many connected subspaces. But must it be the case that any connected metrizable space has uncountably many connected subspaces?
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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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In the context of topological spaces, I see the following major countability properties:  A space is:   "separable" iff it has a countable dense subset "second countable" iff if has a countable basis "first countable" iff the neighbourhood system of every point has a countable local basis.   (Definitions taken from Counterexamples in Topology by Steen and Seebach, 2nd ed. 1978 -- there may be differences in wording from other sources.)  The question I have is: "separable" into what, exactly? By which I mean to say: what is the thinking behind calling such a condition "separable"?  It appears not to be related to the concept of "separation axioms", which do immediately and obviously invoke an intuitive notation of separation, neither does it seem to have anything to do with "separated sets".  (Anyone using the spelling "seperable" or "seperated" will be immediately downvoted. :-) )
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I am self-studying topology from Munkres.  One exercise asks, in part, to show that the spaces $(0,1)$ and $(0,1]$ are not homeomorphic.  An apparent solution is as follows: If you remove a point, $x$, from $(0,1)$, you get a disconnected space; however, you can remove the point $\{1\}$ from the space $(0,1]$ and the space will still be connected.  This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them).   So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure?  Sorry for missing something obvious.