Given two disjoint closed sets, $F_1$ and $F_2$ in $\mathbb{R}$, prove that there exist disjoint open sets $G_1$ and $G_2$ such that $F_1 \subset G_1$ and $F_2 \subset G_2$    The answer is not trivial in that the distance between $F_1$ and $F_2$ could be going towards zero.
123765
Theorem:   Let $S_0, S_1 \subseteq \mathbb{R}^n$ be disjoint and closed.   Then there exist disjoint open sets   $T_0, T_1 \subseteq \mathbb{R}^n$ such that   $S_0 \subset T_0, S_1 \subset T_1$.   How to prove this theorem without Urysohn functions? I know that $S_0^c$ contains $S_1$ and vise versa, but I don't know how to proceed from here.
107553
Let A be a subset of the positive real numbers. It is required to prove  that there exists a metric space whose non-zero distances are exactly the set A. Any suggestions and hints on how to proceed will be highly appreciated. Please try not to post the exact solution.
28137
I would like to prove the following statement.  Let A and B be nonempty disjoint subsets of $\mathbb{R}$ where A is closed and B is compact. Then there exists $a \in A$ and $b \in B$ such that $inf${$|a-b|$:$a \in A, b \in B$}=$|a-b|$    I read a similar question above where the concept of "limit point" and others were used, but my class (Introduction to Analysis) has not gone that far; most of all, I would like to prove the statement from the knowledge that the class covered.  So far, I have learned definition of limit of a sequence, Bolzano-Weierstrass Theorem. Also, I just learned (and proved) that there exists a sequence ${a_n}$ in E such that $\lim{a_n} = \inf{E}$ if E is a nonempty subset of $\mathbb{R}$ (whether E is bounded below or not due to the extended real numbers).   From the things that I learned, first, I constructed a set $E$={$|a-b|$:$a\in{A}, b\in{B}$} and I know that there exists a sequence ${x_{n}}$ in E such that $\lim{x_{n}}=infE$. Since ${x_{n}} \in{E}, x_{n}=a_{n}-b_{n}$ where $a_{n} \in A, b_{n} \in B$. Let $infE = \beta$. Then, this leads to $\lim{(a_{n}-b_{n})}=\beta$. Then, for any $\epsilon$>0, there exists $N \in$ $\mathbb{N}$ such that $|a_{n}-b_{n}-\beta|&lt;\epsilon$ for all n $\ge$ N.   Then, I could not proceed. I should use conditions that A and B are disjoint, A is closed, and B is compact. Could anyone help me proceed? I had a feeling that using contradiction might work from the link above, but it is quite tough. Since I am a novice (just started taking Introduction to Analysis), I would very much appreciate it if you could help me prove this at the beginner level.
76505
Let $X$ be a topological space. Let $A$ and $B$ be two subsets of $X$ such that: $A$ and $B$ are closed, $A \cup B$ is connected, $A \cap B$ is connected     Prove that $A$ and $B$ are connected.   Thoughts:  Suppose that $A$ is disconnected. Since $A$ is closed, there are two non-empty, disjoint, closed sets $E$ and $F$ in $X$ such that $A = E \cup F$. Then I attempted many things but all were not really useful.  Please help. Thanks.
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I'm trying to solve this question:  Let $$M=\{f\in C([0,1])\ | \ f(1)=0\}.  $$  Is $M$ closed in $(C[0,1]),d_1)$?  At first, I assumed that $M$ was closed, so I tried to prove that if there is a sequence $f_1,f_2,f_3,...\in M,$and $\ f_n\rightarrow f\,$ with respect to $d_1$, then $f \in M$.    However, I got stuck and checked out the answers, and the answer I was given was just a counterexample of a sequence $(f_n)_n$ where $f_n \in M,\ f_n\rightarrow f$ wrt $d_1$, and $f\notin M$.    I understand that it's a lot easier to give a counterexample, but unfortunately I suck at analysis and cannot tell whether a set is closed or not upon inspection or come up with a counterexample right away.  Are there other ways of proof, like proof by contradiction, to show that this set is not closed?
22049
I think that the question is clear from the title:     Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$?   My feeling is that the definition of any two such sets will be circular (we can't define $B$ till we define $A$ and we can't define $A$ till we define $B$), and hence will be excluded by some axiom.  On the other hand, I can't see an inherent contradiction, though I wouldn't be surprised if some form of Russell's paradox pops up, pushed behind one more set of braces.
53045
Is there a way to prove this using sequential compactness instead of open cover definitions?  My first gut reaction was that the fact was obvious since we can show that the closed subset $[a,b]$ is compact in $\mathbb R$.
1030
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?  (By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)  Thanks!
50504
Let $A, B$ be subsets of a metric space $X$. If $A$ is compact and $B$ is closed, show that the distance between $A$ and $B$ is achieved.   Attempt at a proof:    Let $A$ be compact and $B$ be closed. Let $m=d(A,B)=\inf_{b\in B} d(A,B)$. Then, there are two possibilities: (a) $\exists b\in B$, $d(a,b)=m$. If this is the case, we're done. (b) $\forall b\in B$, $d(a,b)&gt;m$. In this case,     there exists a sequence $\{b_n\}\subseteq B:$ $d(a,b_n)\rightarrow m$ as $n\rightarrow\infty$ by definition of infinum. Then there exists a subsequence $\{b_{n_k}\}$: $d(b_{n_1},a)&gt;d(b_{n_2},a)&gt;...$ which is monotonic decreasing. Then note that $d(b_{n_k},a)&lt;d(b_{n_1},a)&lt;\infty$. So it's a bounded sequence. Now I want to show that it has a convergent subsequence that converges to $b\in B$ and then I want to do the same for $A$. And then finally to show that $d(a,b)=m$ in fact. Any clues to how to get there?
74509
Wouldn't these two sets have the same cardinality?  If so, I am unsure of how to prove that two sets with the same cardinality have a bijective correspondence.
23729
I have a feeling they are neither closed nor open as the $\mathbb{R} \setminus \mathbb{Q}$ cannot be open or closed either...
98382
Any help with this problem is appreciated. It comes up in the context of Lusin's theorem, where it was assumed to be true.     Suppose $E$ is a closed subset of $\mathbb{R}$ and $f:E\rightarrow \mathbb{R}$ is continuous. There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)=g(x)$ for $x \in E$ and $g$ satisfies $$\sup_{x \in \mathbb{R}} |g(x)| \leq \sup_{x \in E} |f(x)|\;.$$
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I'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.  "Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."  Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.  Thanks.
88541
Let $G$ be the Cantor set. It is well known that:   $G$ is perfect and hence closed. $G$ has the cardinality of the continuum. $G$ has measure zero. For any set $S \subset \mathbb{R}$ (I will not keep writing that we are in $\mathbb{R}$) we have $S \text{ is closed} \Leftrightarrow S^c \text{ is open}$. Any open set $O$ can be written as a --- in fact unique --- countable union of disjoint open intervals.   $G^c$ can thus be written as a countable union of disjoint open intervals. We now imagine this union as being superposed on the real line graphically as follows:  R: &lt;&lt;&lt;----(....)---(..)--(.)---------(...)---&gt;&gt;&gt;   where the (...) represents the open disjoint intervals (of differing size) composing $G^c$, and the ---  represents the remaining non-covered real numbers (that is those in $G$). Now we can cover $\mathbb{R}$ in its entirety by "collecting" the --- into disjoint closed intervals. Any one of these closed intervals might of course consist of only a single element. We now obtain:  R: &lt;&lt;&lt;[--](....)[-](..)[](.)[-------](...)[-]&gt;&gt;&gt;   Take the union of these disjoint closed intervals. This must be $(G^c)^c = G$. Now it is not hard to imagine a mapping from the (...)'s to the [...]'s. Just take the next (...) in line for each [...] (and do some trivial fixing at the ends). Therefore we have written $G$ as a countable union of disjoint closed sets. However, $G$ has measure zero and therefore cannot contain any closed sets other than the single element type. Hence $G$ is countable. Contradiction.  Where do I go wrong?
109363
I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.
106871
Following a reference from "General Topology" by Stephen Willard   Well since the difference between an open set and a closed set is an open set and since the union of open sets is an open set, it is clearly that $S$ and $T$ are open; but unfortunately I can't prove that $S$ and $T$ are disjoint open set containing $A$ and $B$. Anyway it seems to me that $U=\bigcup_{i=1}^nU_i$ and $V=\bigcup_{i=1}^nV_i$ are two disjoint open sets that containing $A$ and $B$ and so I ask: why define the set $S$ and $T$? Could someone help me?
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?