In principals of mathematical analysis on page 32, Rudin defines (definition 2.18e-f): a point $p$ is an interior point of $E$ of there is a neighborhood $N$ of $p$ such that $N\subset E$, $E$ is open if every point of $E$ is an interior point of $E$  By this definition, shouldn't $\mathbb{Z}$ be open? Because $\forall n \in \mathbb{Z} \exists m \in \mathbb{Z}s.t. |n-m|&gt;0$ and so $m$ is in a neighborhood of $n$ and in $\mathbb{Z}$
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Baby Rudin gives the example of the set of all integers being not open if it is a subset of $\mathbb{R}^2$.  If we consider the set of integers in $\mathbb{R}$, is this set also not open? I can find a neighbourhood which will contain any point, $p$, however is it a requirement that a neighbourhood contains more than one point?  I'm trying to understand this fully and have searched through the various posts that have a slight relation and can not find out specifically how these take interior and isolated points into account and how these relate to openess.
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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!
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Previously, someone posted his/her proof of "every neighborhood is open" . I am focusing on the same statement, but it seems that my approach is different from his/hers. Could someone comment on my own proof? Here we go:  Let $X$ be a metric space and $p\in X$. Fix $r&gt;0$. Define a neighborhood around $p$ as $N_r (p)=\{q\in X| d(p, q)&lt;r\}$. Now, take $w = \frac{r}{2}$. Then, $N_w (p)\subset N_r(p)$. In other words, every point of $N_r(p)$ is interior. Therefore, $N_r (p)$ is open. Since $r$ is arbitrary, this proves every neighborhood is open.  Update:  Under my definition of "neighborhood," can I prove in this way instead: For all $m\in N_r(p)$, take $w=\frac{r−d(m,p)}{2}$. Then, $N_w(m)\subset N_r(p)$.   Update2  From the hints given in comments, I solve it:  Assume $m\in N_r(p)$ and $q\in N_w(m)$. Take $w=r-d(m, p)$. Then, $d(p, q)\leq d(p, m) + d(m, q)&lt; d(p, m) + r-d(m,p) =r$.   Thanks to @JoséCarlosSantos and @fleablood !
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I have been stuck on this problem :   Prove :If $A$ is Open , then $A\cap \overline B \subset \overline {A \cap B}$.I have tried to use the previously proven fact that $\overline {A \cap B}\subset \overline{A} \cap \overline{B}$ , but I cannot seem to get anywhere ...In particular, I do not see how A being open is relevant here..........
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Question: Why is the topological space $\mathbb{R}^\infty$ defined to be the subset of $\prod_{i=1}^\infty \mathbb{R}_i$ consisting of sequences $(a_i)_{i=1} ^{\infty}$ such at most finitely many $a_i\neq 0$? Why does one insist on the condition that $a_i\neq0$ for at most finitely many $i$?
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Why is the set of $n\times n$ real, non-singular matrices an  open subset of the set of all $n\times n$ real matrices? I don't quite understand what "open" means in this context.  Thank you.
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Suppose $[\mathbb{Z}_p:G] = n &lt;\infty$. Write $n = p^km$ with $p\nmid m$. The idea is to show that $p^k\mathbb{Z}_p = n\mathbb{Z}_p \subseteq G$, after which I am done, since for any $x\in G$ we have that the open ball $x + p^k\mathbb{Z}_p\subseteq G$, which makes $G$ open.  Unfortunately I am stuck at how to prove that $n\mathbb{Z}_p\subseteq G$. Could someone lend a hand with a hint?
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I am going through Rudin's Principles of Mathematical Analysis in preparation for the masters exam, and I am seeking clarification on a corollary.  Theorem 2.34 states that compact sets in metric spaces are closed. Theorem 2.35 states that closed subsets of compact spaces are compact. As a corollary, Rudin then states that if $L$ is closed and $K$ is compact, then their intersection $L \cap K$ is compact, citing 2.34 and 2.24(b) (intersections of closed sets are closed) to argue that $L \cap K$ is closed, and then using 2.35 to show that $L \cap K$ is compact as a closed subset of a compact set.  Am I correct in believing that this corollary holds for metric spaces, and not in general topological spaces?
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What does $\mathbb{R}\setminus S$ mean? I am not getting it what it actually means. I have found it manywhere in real-analysis like in the definition of boundary points of a set. Can anyone tell me what it means really?
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The following proof is based on Rudin's Principles, chapter 2, exercise 26. I wanted to confirm that the axiom of (countable) choice is used in the line in bold, and was curious to know whether this proof can be modified so that it isn't required, and if not, whether there is an entirely different proof which doesn't require choice.  Let $X$ be a metric space in which every infinite subset has a limit point. By previous exercises, we know that $X$ must have a countable base. So every open cover of $X$ has a countable subcover $\{ G_{n} \}$. Assume that $\{ G_{n} \}$ has no finite subcover. Construct a set $E$, containing one element from each set $F_n = (\bigcup_{i=1}^{n}G_i)^c$. Then $E$ is infinite, since every point in it is in some $G_m$, so won't be in $F_{m}$. But then $E$ must have a limit point $y$, which is in contradiction with the fact that $y$ must be in some $G_n$, which is open. Thus $\{ G_{n} \}$ must have a finite subcover.
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I would be very glad if you helped me solving this problem. It's about topology. Okay let's begin :    Let $A$ be a non-empty open subset of $\mathbb R$. Let $x_0 \in A$ and $B = [x_0,\infty) ∩ A^c$ (complement of $A$)  First I should prove that $B$ is a non-empty, closed set. Then I should prove that :     $B$ is bounded below.   $\inf B &gt; x_0$   $\inf B$ doesn't belong to $A$.   $[x_0, \inf B) \subseteq A$.    Then I should explain how to deﬁne a number $\beta$ so that $(\beta,\inf B) \subseteq  A$, where $\beta$ doesn't belong to $A$.    I really can't get up with proofs for these question. Can you please help me solving this or at least give me some hint or sketch? I would be very grateful. Thank you :)
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In the middle of another proof (Theorem 3.4, p. 60) in his Functional Analysis book, Rudin says that  "every nonconstant linear functional on $X$ (topological vector space) is an open mapping."   Is this obvious? Could someone please show me how the proof goes?
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Question: Let If $P\subseteq \mathbb{R}^n$ be a convex set. show that $int(P)$ is a convex set.  I know that a point $x$ is said to be the interior point of the set $P$ if there is an open ball centred at $x$ that is contained entirely in $P$. The set of all interior points of $P$ is denoted by $int(P)$.  Also a set is convex if $x,y \in P$ $\implies$ $(tx+(1-t)y) \in P$ for all $t \in (0,1)$.  How to go about the above proof?
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Prove that for any $x_0 \in X$ and any $r&gt;0$, the open ball $B_r(x_o)$ is open.   My attempt: Let $y\in B_r(x_0)$. By definition, $d(y,x_0)&lt;r$. I want to show there exists an $r_1\in\mathbb{R^+}$ s.t. $B_{r_{1}}(y)\subseteq B_r(x_0)$. Let $a\in B_{r_{1}}(y)$. Then, $d(a,y)&lt;r_1$. For $a\in B_{r}(x_0)$, $d(a,x_0)&lt;r$. I want to show $d(a,y)&lt;r_1$ implies $d(a,x_0)&lt;r$. By triangle inequality, $d(a,y)\leq d(a,x_0) + d(y,x_0) \rightarrow$  $d(a,y)&lt;r_1\leq d(a,x_0)+d(y,x_0)&lt;2r...$  I'm a little stuck after this point.
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$X=(-\infty,0]\cup\left\{{1\over n}:n\in\mathbb N\right\}$ 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?
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Let $X = \{(x,y) \in \mathbb{R^2}: x &gt; 0, y = sin(1/x)\}$. Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?  I just want to make sure I am thinking about these problems in a reasonable manner. I am also trying to apply this to other concepts I have learned.  Is $X$ open?  X is open if every $x \in X$ is an interior point. Take an arbitrary $p \in X$ so that $p = (p_x, p_y)$ then $p_x &gt; 0$ and $p_y = sin(1/p_x)$. Define an open ball $B_r(p)$, and take $q \in B_r(p)$ be an arbitrary point. Is $q \in X$? Is it possible for $q$ to not be in X? Since $y = sin(1/x)$ is defined and is continuous for all $x \neq 0$,then there will always exist a $q_y = sin(1/q_x)$, provided $q_x \neq 0$. How do I know $q_x \neq 0$? Otherwise, I propose that X is open.   Alternatively, X is open if whenever $f: \mathbb{R^2 \to R}$ is continuous and $U \subseteq R$ so that if $U$ is open, then $f^{-1}(U)$ is also open. Could I construct such a $U$ to prove that it's open? My only concern is that if $U = (-1,1)$, then $f^{-1}(U) = (-\infty, \infty) = \mathbb{R}$ which is not open nor closed.   Is $X$ closed?   I am not sure about this one. I want to say that is not closed, since $X^c$ is not open. It is not open because $(0,0)$ defines a boundary point which lies in the set $X^c$.  Thanks!
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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?