If A and B are compact set, then A∩B is compact set.  How to prove this problem?.  I know compact, but not use to this problem...
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Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.  I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.  Thank you for the help
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I found the following question about compactness:     A set $S\subseteq \mathbb{R}^n$ has the following property:   if a sequence of $S$ has exactly one accumulation point in $S$, then it converges in $S$. Prove that $S$ is compact.    We know that in $\mathbb{R}^n$ compact sets correspond to closed and bounded sets.   (1) I am trying to prove that $S$ is closed. The condition in the question says if a sequence of $S$ has an accumulation point in $S$ then it converges in $S$. I think instead the condition should be if a sequence of $S$ has an accumulation point( in $\mathbb{R}^n$, not necessarily in $S$ )then it converges in $S$.  Am I correct?  (2) How to prove that $S$ is bounded?  P.S. I have made an edit (there is exactly one accumulation point).  Source of the question:  of this problem set.
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Hi I just want to find out if the following is a acceptable proof for the proposition:   "Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."  Proof:  Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$,  $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.   W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.   $\square$
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Claim. Let $a$ and $b$ be real numbers. If $a\geq b$  then $(a,b), [a,b)$ and $(a,b]$ are empty set.   My is question is that why?
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I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
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While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)?   What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology?
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Let S be an infinite subset of H  Assume that no point of H were a limit point of S.  Since H is compact, there exist a collection of open sets V={Vq: q∈H }  Vq∩S is finite (Is this true? If yes why?)  Since S is assumed to be infinite  ⋃Vq ⊃ S cannot cover S  Since S⊂H  ⋃Vq ⊃ H (cannot cover H)  This contradict compactness of H.  Hence every infinite subset of H has a limit point (Why do we conclude this?)
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Hello I am taking a Discrete Mathematics course and having some trouble with this question about sets:  Under what conditions is $A- B = B- A$  Diagram:   Diagram:   Maybe I'm understanding incorrectly, but how can A-B and B-A be equal if they contain elements that are not in each other?
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If $(X,\mathcal{T})$ is a connected space, and $Y$ a connected subset, and $X\setminus Y=A\cup B$ for separated sets $A$ and $B$, then why is $A\cup Y$ connected as well?  Thank you kindly.
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Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?
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Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$?
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Statement: $[a, b] \cap \mathbb{Q}$ in $\mathbb{Q}$ is not compact. Thus the interior of all compact subsets of $\mathbb{Q}$ is $\emptyset$.   I am trying to understand the first sentence. I read that a closed subspace of a compact space is compact, so for example, consider the unit interval $[0, 1]$ which is a compact space. Take a closed subspace $[0, 1] \cap \mathbb{Q}$ of $[0, 1]$. This set is closed since it just consists of all the rational numbers in between $0$ and $1$, including $0$ and $1$. So it is a closed subspace of a compact space. But why isn't this compact?
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Let $A , B\subseteq\mathbb{R}$. If $A$ is closed and $B$ is compact, is $A\cdot B=\{a. b: a\in A,b\in B\}$ closed?  If $p$ is an adherent point in $A\cdot B$, exists a sequence $p_{n}=\alpha_{n} \beta_{n}$, $\alpha_{n} \in A$ and $\beta_{n}\in B$, such that $p_{n}\rightarrow p$. If $B$ is compact, $\beta_{n}\rightarrow b \in B$.
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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.
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Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$  I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise. Edit: Thanks for the input. FYI, I need to prove this using predicate logic.
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Let $B$ be a compact subset of a metric space $X$. Let $x \in X \setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x \in U$ and $B \subseteq V$.   Is this an existence proof? If yes, do I get to define $U,V$?
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Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$.  What is the correct and mathematically strict way to prove the above? (slightly different than  )