every infinite set of points in M has a accumulation point in M => M is compact? The answer in the textbook says no. But I have a proof for it, so wanna know where the mistake is.  proof :  supposed M not compact.  Let { Ui } be a open sets covering of M, which doesn't have a finite subcovering.  Let {Uj} be a minimal subset of {Ui} such that {Uj} is still a covering of M. (in other word : when you remove any element from {Uj}, the remained {Uj} don't cover M any more.)  { xj : xj is an element of Uj , xj is different from xk when j is not k} You take a point xj from each Uj, which is different from any other points from other Ujs( in other word : which is not contained in any other Uj).  it's possible because {Uj} is minimal.  This set of xj has a accumulation point p in M.   Since {Uj} cover M, there exist a Uj which contains p.  because p is a accumulation point of xj and Uj contains p, Uj contains infinate many xj. This is a contradiction to the choice of xj. (since each xj is contained in only one Uj)  Hope to find my mistake. thank you very much.  --------------replies to comments-------------  (I can't post comment so I have to write it here)  @Max   My question is not that, is actually the reverse implication.
What's going on with "compact implies sequentially compact"? I've seen both counterexamples and proofs to &quot;compact implies sequentially compact&quot;, and I'm not sure what's going on. Apparently there are compact spaces which are not sequentially compact; quick googling and wikipedia checks will turn up examples floating around; they tend to be variants of $[0,1]^{[0,1]}$ with the product topology.  Here's a demonstration(?):  $[0,1]^{[0,1]}$ is compact by Tychonoff's theorem.  So, we demonstrate failure of sequential compactness. Choose a unique binary representation for each $x\in [0,1]$. For each $n\in\mathbb{N}$, let $f_n : [0,1]\to[0,1]$ (an element of $[0,1]^{[0,1]}$) be the function which maps each $x$ to its $n$-th place digit in its binary expansion. Let $f_{n_k}$ be a subsequence of this sequence. Let $x'\in [0,1]$ be such that the $n_{2m}$-th digit is $0$ and the $n_{2m+1}$-th digit is $1$, for all $m\in\mathbb{N}$. Then $f_{n_k} (x')$ does not converge (it alternates between $0$ and $1$), and hence $f_{n_k}$ cannot converge. Thus we have found a sequence in $[0,1]^{[0,1]}$ without any convergent subsequence and so it is not sequentially compact.  (Aside: this is apparently based off of an example in Steen's Counterexamples in Topology, according to  ) Nevertheless, there's also some proofs(?) floating around that compactness of a space implies sequential compactness, along these lines (this proof(?) is of the contrapositive):  Suppose $X$ is not sequentially compact. By definition, this means there is some sequence $(x_n)$ over $X$ with no convergent subsequence. If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption.  (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.) Thus, for each $x \in X$ we can select an open set $U_x$ such that $x\in U_x$ but with $x_n \in U_x$ for only finitely many $n$. The collection $\mathcal{U}=\{U_x : x\in X\}$ is clearly an open cover of $X$. If $\mathcal{U}$ had a finite subcover $\{U_1, \dots, U_k\}$ then the union $U_1 \cup \cdots \cup U_k$ would contain all of $X$ but only contain $x_n$ for finitely many $n$, which is impossible. Thus $X$ is not compact, since we have found an open cover without a finite subcover.  (Aside: this proof is essentially the same as the proof appearing in Rudin's Principles of Mathematical Analysis for Thm 2.37 that infinite subsets of compact spaces have limit points.) So, what's going on here?  It can't be that what both the counterexample and the proof are telling us is correct.  Is there some subtle (or more embarassingly for me -- glaring) flaw in the proof?
what is wrong with this??? please,can anyone help?  \begin{equation} \label{probabilityy} P^{t}({E_{0})=P_{0}+(1-P_{0})P_{1}+ (1-P_{0})(1-P_{1})P_{2}+.......+P_{n}\Pi_{k_{0}}^{n-1}(1-P_{k}) \end{equation} Allow deleting a tag wiki or excerpt Sometimes a tag wiki excerpt or body ends up with unacceptable or useless content: plagiarized text that must be removed, or a totally unhelpful dictionary definition. (For example:  — and I've seen this situation many times before.)  The only way to get rid of that content is to submit replacement text that meets the minimum length requirement. This is not always easy as writing a decent wiki, and especially a decent excerpt, requires subject knowledge.  Currently the users or even moderators who are alerted to unacceptable content in a tag wiki cannot remove that content without providing their own, which they are not always in a position to do. 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What did you say?  Did you offend someone?  We encourage people to come forward.  We don't have to explain why someone took our money.     I think the "No one ever [...]" claim is literally false, and actually obviously so. So, I understand it to mean:          The President also knows that we have to stop blaming victims for these crimes.  No one in their right mind ever asks the person who got robbed at gunpoint in the street -- why were you there, what were you doing, what were you wearing?  What did you say?  Did you offend someone?  We encourage people to come forward.  We don't have to explain why someone took our money.      Supposing that I am right, what is this kind of rhetoric called? A real nxn matrix always has a real eigenvalue when n is odd what do it mean by "A real nxn matrix always has a real eigenvalue when n is odd?" 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