$\varnothing\neq A \in 2^\mathbb R$, $A^s$ is isolated points of $A$. Is $A^s$ closed set?  notation: x is isolated point means $\exists r&gt;0 [N(x;r)\cap A=\{x\}]$  I would like to prove but I have no idea. I'm a uni student studying the very beginning of topology. Could you help me.
121620
In metric space,  $(X,d)$  I know limit points of some set(S maybe) is closed in X  Then is set of isolated points of S is closed?   Want to make sure if closure of isolated points of S doesn't intersect with limit points of S  But i think i got it and i think i should leave it for people like me
107715
My question is  Let $A: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation, how can one prove that $\{Ay|y\geq 0\}$ is a closed set?  Here $y\geq 0$ means each component of $y$ $\geq 0$. I can simplify the question in two ways  Since the image of $A$ is a closed linear subspace of $\mathbb{R}^m$, by replacing $\mathbb{R}^m$ with this linear subspace, we may assume A is surjective.  After a change of basis of the codomain and a reordering of the basis of the domain, we may assume $A$ is of the form $[I|B]$   Any help or hints are appreciated, thank you.  Ideas: Let $v_i=Ae_i$, then my original statement is equivalent to saying the set $\{\sum a_iv_i |a_i\geq 0\}$ is a closed set. So I should now prove the following:  For any $v_1,...,v_k \in \mathbb{R}^n$, the set $S=\{\sum a_iv_i |a_i\geq 0\}$ is a closed subset of $\mathbb{R}^n$.   I suspect that the set $S$ is an intersection of half spaces, from this the closedness will be obvious. Equivalently, I suspect that for each $x\notin S$, there exists a half space $H$ (cut by a codimension 1 linear subspace)s.t. $S\subset H$. Is this statement true? I guess this should be true after drawing some examples.  New ideas: I am proving that the set $\{\sum a_iv_i|a_i\geq 0 \}$ is closed, and this is easy if all the $v_i$'s are indeed independent, so I would want to prove the following:  Let $S=\{$linearly independent subsets of $\{v_i\}\}$, then $\{\sum a_iv_i|a_i\geq 0\}=\cup_{T\subset S}\{$non negative linear combinations of $T\}$  And this union is closed since it is a finite union of closed set. Equivalently, let $x=\sum a_iv_i,a_i\geq 0$, I want to write $x=\sum b_ie_i$, where $b_i\geq 0$ and $\{e_i\}\subset \{v_i\}$ is linearly independent. My proof is as follows: I can replace $\{v_i\}$ with a smaller subset of itself s.t. $x$ cannot be expressed by non negative linear combinations of any strictly smaller subset of it. Now I claim that this set is linearly independent. Suppose not, then we can write $c_1v_1+c_2v_2+\cdots + c_nv_n=0$, with some $c_i$'s are positive(since we can multiply the whole expression by $-1$) and I assume that $c_i&gt;0$ for $i=1,2,\dots ,k$ and $c_i \leq 0$ for $i&gt;k$ Now I assume $b=a_1/c_1=\min _{i=1,2,\dots ,k}{a_i/c_i}$, Then I have $$x=\sum a_iv_i-b(\sum c_iv_i)=                      (a_2-bc_2)v_2     +(a_3-bc_3)+\cdots + (a_n-bc_n)v_n$$ Which is a non negative linear combinations of fewer vectors, contradicts to my assumption and hence the minimal set $\{v_i\}$ is linearly independent.
111671
$X$ is a topological space with no special properties.  If a space $X$ has no isolated points, then show the same for any dense subset of   $X$.  Thanks a lot.
25276
The question is:     Show that if $A⊆\Bbb R^n$ is finite and $n≥2$, then the set $\Bbb R^n\backslash A$ is connected in $(\Bbb R^n,τ_d)$ where $τ_d$ is the topology induced by the Euclidean distance in $\Bbb R^n$.   My attempt:  Denote $A^C=\Bbb R^n\backslash A$ and suppose $A=\{a_1,…,a_n\}$. Assume by contradiction that $A^C$ is not connected, then there exists $O_1,O_2∈τ_{A^C }$ st. $A^C=O_1⋃O_2,O_1⋂O_2=∅$. Here $\tau_{A^C}$ is a subspace topology.  Intuitively, part of $A$, denoted by $A_1$ is "enclosed" in $O_1$ and the rest, denoted by $A_2$ are "enclosed" in $O_2$. Then $G_1=O_1\bigcup A_1$ is an open set in $\tau_d$, and $G_2=O_2\bigcup A_2$ is also an open set in $\tau_d$, and $\Bbb R^n=G_1\bigcup G_2$. However, $G_1, G_2$are disjoint, contradicting with the fact that $\Bbb R^n$ is connected.  However I got stuck when trying to implement this idea. Can anyone provide some help? Thank you!
80961
I don't understand the following proof of that every interval is connected in $\mathbb{R}$.     Let $Y$ be an interval in $\mathbb{R}$ and suppose that $Y$ is not connected.      Then $Y=A\cup B$, where $A,B\subseteq Y$ are open in $Y$, $A,B\neq\emptyset$ and $A\cap B=\emptyset$. Let $a\in A$ and $b\in B$. Without loss of generality, $a&lt;b$.       Let $\alpha=\sup\{x\in\mathbb{R}:[a,x)\cap Y\subseteq A\}$.      Then $\alpha\le b$ and $\alpha\in Y$. It is clear that $\alpha\in Cl_{Y}(A)$, and $A$ is closed in $Y$, then $\alpha\in A$. Since $A$ is open in $Y$, $Y$ is an interval, $b\in Y\setminus A$ and $\alpha &lt;b$, then there exists $r&gt;0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$. We can conclude that $[a,\alpha+r)\cap Y\subseteq A$, which is a contradiction.   I don't understand why $\alpha\le b$ and $\alpha\in Y$. Can anyone explain this to me?  Thanks.
67644
So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n &gt; \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily).
97600
My question is about the following claim:     For $n \geq 2$, let $A\subset \mathbb R^n$ be a non-empty, open, bounded set. Assume $A$ and its complement are connected and $\text{int}(\text{cl}(A)) = A$. Then $\partial A$ is connected.   Without the assumption $\text{int}(\text{cl}(A)) = A$, the statement is false (just take any open set and remove an interior point). This condition is not necessary but I think it is sufficient to get the claim.  This seems a trivial matter, but I cant's find a proof using only basic topological tools. Does anyone know something about this ? Thank you very much.
105612
This is a Theorem in Conway's book     I don't understand in the third paragraph why there are open sets $\{U_n:n\geq1\}$ in $(ball X^*,wk^*)$ such that $0\in U_n$ and $\bigcap^{\infty}_{n=1}U_n=\{0\}$. By the definition of the relative weak-star topology on ball $X^*$, for each $n$, there exists a finite set $F_n$ contained in $X$ such that $$\{x^*\in\text{ball }X^*:|\langle x,x^*\rangle|&lt;1 \text{ for all }x\in F_n\}\subseteq U_n.$$  Can someone explain this for me？ Thank you so much!!
102487
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!
12106
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?
65959
Let $S$ and $V$ be the following subsets of $\mathbb{R}^2$.  $$ S = \left\{ \ x \times \sin \frac{1}{x} \ \colon  \ 0 &lt; x \leq 1 \ \right\}, \qquad V = \left\{ \ 0 \times y \ \colon \ -1 \leq y \leq 1 \ \right\} = \{ 0 \} \times [-1, 1]. $$ Then how to show rigorously that the closure $\overline{S}$ of $S$ is given by  $$ \overline{S} = S \cup V? $$  By rigorously, I mean using the smme kind of argument as is used in real analysis or topology, such as in (most of) the proofs in Rudin or Munkres.   Intuitively, it is clear that as $x \to 0+0$, then $\sin \frac{1}{x}$ will fluctuate its full range of $[-1, 1]$, and for any point $0 \times y$, where $y \in [-1, 1]$, there will be points of $S$ arbitrarily close to it.   But I'm unable to rigorously show this.
60920
How to prove that a compact set $K$ in a Hausdorff topological space $\mathbb{X}$ is closed? I seek a proof that is as self contained as possible.  Thank you.
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?
11662
In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?  I tried by assuming a set is equal to its interior points and contains its limit points.  A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?  Anyway, it must then be unbounded.  If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$.
87173
Let $X$ be infinite set and $a \in X $. Let $$\mathscr{F} = \{ A \subset X: |A| \; \; finite \; \;, a \notin A \}$$  Then $\mathscr{F}$ is the set all of all closed sets for some topology $\mathcal{T} $ on $X$. Also, any two continuous maps $f,g: (X, \mathscr{T}) \to Y$ ( $Y$ is metric space) have the property that if $f(b) = g(b)$ for all $b \neq a $, then $f(a) = g(a)$.  In my notes my professor posted this theorem without proof, so I was wondering what is best way to proof this theorem.
73771
A point $p$ in a metric space $X$ is a boundary point of the set $A$, if any neighbourhood of $p$ has points of both $A$ and $X-A$.Prove that the set of all boundary points of $A$ is closed.  My attempt: By definition of an open set this means that for every $x$ in the boundary there is an open ball centred at $x$ contained in the boundary. An open ball is a neighbourhood of $x$, which implies it contains points of $A$ and $X - A$, which in turn implies there are points in both $A$ and $X - A$ that are in the boundary of $A$.  If $A$ is open, then pick any such point in $A$ that is also in the boundary. This point cannot be in $X - A$ by definition of set subtraction. Further, because $A$ is open there exists an open ball centred around this point contained in $A$. Again, an open ball is a neighbourhood, which means a neighborhood of this point does not contain points of $X - A$, implying it cannot be in the boundary, a contradiction. If A is closed then $X - A$ is open and a symmetric argument holds. Hence the boundary is closed.  Is my work correct?
116176
Question:  Prove that a finite subset in a metric space is closed.  My proof-sketch: Let $A$ be finite set. Then $A=\{x_1, x_2,\dots, x_n\}.$ We know that $A$ has no limits points. What's next? Definition: Set $E$ is called closed set if $E$ contains all his limits points.  Context: Principles of Mathematical Analysis, Rudin