Let $A = \{(x, y) \in \mathbb{R}^2 \mid xy &gt; 0 \}$. Show that $A$ is a open subset of the plane.  I have that $A = \{(x, y) \in \mathbb{R}^2 \mid x &gt; 0,y&gt;0 \} \cup \{(x, y) \in \mathbb{R}^2 \mid x&lt; 0, y&lt;0 \}$ Letting $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x &gt; 0,y&gt;0 \}$ and $r=\min\{x,y\},$$B((x,y), r)$ I have that $$|x-x'| \leqslant \sqrt{(x-x')^2 + (y-y')^2} = d((x,y), (x',y')) \leqslant r &lt; \frac{r}{2}$$ so $x' &gt;\frac{r}2 &gt;0.$ Similar reasoning can be done for $y'$. Which would show that $B((x,y),r) \subset \{(x, y) \in \mathbb{R}^2 \mid x &gt; 0,y&gt;0 \}$. How should I deal with the set $\{(x, y) \in \mathbb{R}^2 \mid x&lt; 0, y&lt;0 \}$? I cannot seem to show this... If I let $(x,y) \in \{(x, y) \in \mathbb{R}^2 \mid x&lt; 0, y&lt;0 \}$ and again $r = \min\{x,y\}$ and construct a ball $B((x,y), r)$ I can pick $(x',y') \in B((x,y), r)$, but I'm not sure how to proceed from here?
121079
Show that $$B=\{(x,y) \in \Bbb{R^2} \mid \ xy&gt;0 \}$$ is an open set in the Euclidean space $(\Bbb{R^2},d_2),$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.  Defining $A= \{(x,y) \in \Bbb{R^2} \mid x&gt;0,y&gt;0 \}$ and $E=\{  (x,y) \in \Bbb{R^2} \mid x&lt;0,y&lt;0 \}$ one sees that $B= A\cup E$ and thus $B$ will be open if $A$ and $E$ are open. How can I show that $A$ and $E$ are both open? It seems that I would need to find open balls with arbitrary radius $B((x,y), r)$ for both cases that are contained in the set I'm working with?
61381
This is an example from Munkres's Topology (Example 4 in Section 22 titled &quot;The Quotient Topology&quot;, 2nd edition).  Example 4: Let $X$ be the closed unit ball $$\{ x \times y \mid x^2 + y^2 \le 1\}$$ in $\mathbb{R}^2$, and let $X^{\ast}$ be the partition of $X$ consisting of all the one-point sets $\{ x \times y \}$ for which $x^2 + y^2 &lt; 1$, along with the set $S^1= \{ x \times y \mid x^2 + y^2 = 1 \}$. Typical saturated open sets in $X$ are pictured by the shaded regions in the figure below. One can show that $X^{\ast}$ is homeomorphic with the subspace of $\mathbb{R}^3$ called the unit 2-sphere, defined by $$S^2 = \{ x \times y \times z \mid x^2 + y^2 + z^2 =1 \}.$$  I am confused about two points in the example.  Problem 1: About the saturated open sets in $X$. In the example, the typical ones are pictured by the shaded regions ($U, V$) in the figure. However, I am not sure whether the boundaries (visually, by picture) of $U$ and $V$ are included. Particularly, there are two boundaries for $U$. Are they both contained in $U$? And why? Problem 2: How to show that $X^{\ast}$ is homeomorphic with $S^2$? What is the mapping between them?  The following is my understanding:  Solution to problem 1: For $V$, the boundary is not included. For $U$, the outer boundary is included, while the inner one not. Is this solution right?
56116
One exercise from a list. I have no idea how to finish it.  Let $I=[c,d]\subset \mathbb{R}$. Let $f:I\to \mathbb{R}$ be continuous at $a\in (c,d)$. Suppose that there exists $L\in \mathbb{R}$ such that $$\lim \frac{f(y_n)-f(x_n)}{y_n-x_n}=L$$ for every pair of sequences $(x_n),(y_n)$ in $I$, with $x_n&lt;a&lt;y_n$ and $\lim x_n=\lim y_n=a$. Prove that $f'(a)$ exists and it is equal to $L$.  Any help? Thanks in advance.
10486
Let $X \subset \mathbb{P}^n$ be a nonempty projective variety. Show that the dimension of the cone $C(X):=\{0\} \cup \{(x_0,...,x_n)\in \mathbb{A}^{n+1}:(x_0:...:x_n)\in X\}$ is dim$X+1$.  I know how to prove dim$C(X) \geq$ dim$X+1$: Let $\varnothing \neq X_0 \subsetneq ... \subsetneq X_n\subset X$ be a chain of irreducible closed subsets in $X$, then $\{0\} \subsetneq C(X_0) \subsetneq ... \subsetneq C(X_n) \subset C(X)$ is a chain of irreducible closed subsets in $C(X)$.   However I cannot prove the other side of the inequality, if I let $\varnothing \neq Y_0 \subsetneq ... \subsetneq Y_m\subset C(X)$ be a chain of irreducible closed subsets in $C(X)$, I do not know how to generate a chain of irreducible closed subsets in $X$. If the $Y_i$ are of form $V(S_i)$, where $S_i$ is a homogenous ideal, then I can take projectivization $\mathbb{P} (Y_i):=\{(x_0:...:x_n)\in \mathbb{P}^{n+}:0 \neq (x_0,...,x_n)\in X\}$ and obtain my conclusion easily. Is there a way to assert this? This is equivalent to the claim that every irreducible component of a cone is also a cone, which seems to have some geometric basis.  I have seen some answers on related questions on this site using the tool of fiber dimension and transcendence degree of the coordinate ring, however I do not want an answer with these advanced techniques, as the notes I am reading does not assume the reader has these knowledge; it would be very nice to see a proof using basic constructions such as cone and projectivization defined above.  Any help is appreciated!
85750
Are there two closed sets $A,B\subset\mathbb{R}^2$ with the following properties?   $A\cap B=\emptyset$ $\forall \epsilon&gt;0$ there exist $a \in A$ and $b\in B$ such that $\|a-b\| &lt; \epsilon$
87752
$X=(-\infty,0]\cup\left\{{1\over n}:n\in\mathbb N\right\}$ with subspace topology. Then   $0$ is an isolated point $(-2,0]$ is an open set $0$ is a limit point of the subset $\left\{{1\over n}:n\in\mathbb N\right\}$ $(-2,0)$ is open set.   $0$ is not an isolated point of $X$ as every nbd of $0$ contains a point of the form $1\over n$ so  $3$ is true and $1$ is false, $2$ is false as $0$ is not interior point, $4$ is true as every point is interior point. Am i right?
97559
how to prove that if $A$ and $B$ are open in $(\mathbb{R},|.|)$ then  $A+B$ is open ?  Where   $A+B=\{a+b\mid a\in A,b\in B\}$   Thank you
107671
One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.
25548
I need to show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric. I know that a set $U$ is open if for $x\in U$ there exists an open ball $B_\epsilon(x)$ such that $B_\epsilon(x)\subset U$ for some $\epsilon &gt;0$. The euclidian metric in $\mathbb{R}^2$ is given by $d_2(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$, so I need to show that for $x\in (a,b)\times (c,d)$ there exists $B_\epsilon(x) = \{y\in\mathbb{R}^2:\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} &lt; \epsilon\}$ such that $B_\epsilon(x)\subset (a,b)\times (c,d)$ for some $\epsilon &gt;0$, but I have no idea how!  Question: How do I show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric?
110211
Is every open subset of $\mathbb{R}$ uncountable?  I was crafting a proof for the theorem that states every open subset of $\mathbb{R}$ can be written as the union of a countable number of disjoint intervals when this question came up.  I feel like the answer is yes, but I'm not sure how to go about proving it or whether there is a crazy construction (like the Cantor Set)  that creates a countable, open subset of $\mathbb{R}$. Any ideas?
65114
Let $ Y = \{ (1,1),(0,0\} \subset \mathbb A_k^2$. Find the $ \mathbb I(Y)$.   As if $f(x,y) \in \mathbb I(Y)$ then as $f(0,0)=0$ hence constant term of $f$ is zero and as $f(1,1)=0$ therefore sum of the coefficients of $f(x,y)$ is zero. But how can we describe this set properly?
39347
Let $X$ be a subset of $\mathbb{R^n}$. The topology on $X$ is induced by the topology of $\mathbb{R^n}$. If there is an homeomorphism from $X$ onto $\mathbb{R}^n$, is it true that $X$ is open in $\mathbb{R}^n$ ? I think yes, but how can it be shown? (of course, $X$ is open in $X$).  If $x \in X$, $H_{n-1}(\mathbb{R}^n-\{f(x)\})= \mathbb{Z}$ so $H_{n-1}(X-\{x\})= \mathbb{Z}$, but I don't know how to continue.  Thanks in advance.
92238
Let $n \in \mathbb N$ and $U\subset \mathbb R^n$ be an open set.Let $K\subset U$ be a compact subset of $U$. Then How can we show that distance between $K$ and $\partial U$ is positive?   Please, someone, give some hints. to solve this.  Thank you.
50191
How to give a direct proof of the following result?     Let $A$ be a subset of $\mathbb{R}$ such that $A$ is both open and closed. Then $A$ is either empty or all of $\mathbb{R}$?   My work:   If $A$ is not empty, let $u \in A$. Then there is some open interval $(a,b)$ such that $$u \in (a,b) \subset A.$$  Now if $\mathbb{R} - A$ is not empty either, then let $v \in \mathbb{R} - A$. Then there is some open interval $(c,d)$ such that  $$v \in (c,d) \subset \mathbb{R} - A.$$  Suppose that $u &lt; v$.   Since $$\emptyset \subset (a,b) \cap (c,d) \subset A \cap (\mathbb{R} - A) = \emptyset,$$ we must have $$(a,b) \cap (c,d) = \emptyset.$$ So we can conlclude that $$b \leq c.$$ But $b \in A$ and $c \in \mathbb{R} - A$. So we must have $$b &lt; c.$$  What next? Can anybody here please help complete the proof from here on?  An edit based on a comment by Marc Paul:   Let us assume that the set $A$ is a non-trivial clopen set in $\mathbb{R}$. Let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ as follows:  $$f(x) \colon= \begin{cases} 1 \ \mbox{ for } \ x \in A; \\ 0 \ \mbox{ for } \ x \in \mathbb{R} - A. \end{cases}$$   Let $V$ be an open set in the range space $\mathbb{R}$. We show that the inverse image set $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   The following cases arise:   If $0, 1 \in V$, then $f^{-1}(V) = \mathbb{R}$.   If $0 \in V$ but $1 \not\in V$, then $f^{-1}(V) = \mathbb{R} - A$.   If $1 \in V$ but $0 \not\in V$, then $f^{-1}(V) = A$.   And, if $0 \not\in V$ and $1 \not\in V$, then $f^{-1}(V) = \emptyset$.   Thus, $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   Hence the function $f$ is continuous.   What next? How does this lead to our desired conclusion?   [Yet another edit, again based on valuable comments from Marc Paul: ]  So if both $A$ and $\mathbb{R} - A$ were non-empty, then let's suppose $a \in A$ and $b \in \mathbb{R} - A$, and we can assume without any loss of generality that $a &lt; b$.   Then $f(a) = 1$ and $f(b) = 0$. So by the intermediate value theorem there is a real number $c \in (a,b)$ such that $f(c) = 1/2$, which is a contradiction because the image set of $f$ does not contain $1/2$.
39625
consider $r&gt;0 , p&gt;1$ and $K \subset B(x_0 , 2r) \subset R^n$ . $K$ compact. Define the sets :  $$A = \{ u \in C^{\infty}_{0} (B(x_0 , 2r)); \textit{ such  that }  \  u=1 \textit{ in a open  neighborhood of K} \}$$  $$ B =  \{  u \in  H^{1,p}_{0} (B(x_0 , 2r)) \cap  C (B(x_0 , 2r))  : u=1 \  on \  K  ,  \  0 \leq u \leq 1\}$$  I want to show that   $$ \displaystyle\inf_{u \in A} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p  \ dx= \displaystyle\inf_{u \in B} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p   \ dx$$  For this i am trying to aproximate (in the sense of the  norm of $H^{1,p}$) the function of B , by function of $A$.  Somenone can give me a hint ?  thanks in advance
58891
Suppose that $S$ is a closed subset of $\mathbb R^d$ and that $f$ is a continuous real-valued function with domain $S$. Show that, for every real number $a$, the set {$x ∈ S | f(x) ≤ a$} is a closed subset of $\mathbb R^d$    I don't know how to start this. Do we use the fact that If $S$ is closed and bounded every sequence in $S$ has a subsequence that converges to a point in $S$?
15936
So I've been trying to prove that a set $\mathbb{R}$ \ $\mathbb{Z}$ is an open set but I don't quite know where to start since the relative complement is still infinite. So theoretically in $\mathbb{R}$ \ $\mathbb{Z}$ one could pick a real number as close to an integer as needed but you could of course never pick the integer itself. So my idea was that $a-\epsilon&lt;a$ for any $\epsilon&gt;0$ with $a\in \mathbb{Z}$ but I have a feeling I'm doing it wrong. Any ideas or hints on how to solve this properly?