Critique this proof on compactness. Problem: Prove or disprove, the metric space $X$ containing infinitely many points with the discrete metric is compact. Write a proof in the language of sequences and covers

Proof: Take $(1/n) \to 0$. Every term of $(1/n)$ is distinct, and $d(x_n, x) \nrightarrow  0$ as $n \to \infty$. Since every subsequence must also have distinct terms, then the subsequence must also satisify $d(x_{n_k}, x) \nrightarrow  0$

 A: Some notes before Brian M. Scott gives the complete run-down.
Open cover proof


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*If $X$ is uncountable, then $\{ N_1 ( x_i) \}_{i > 1}$ will not cover $X$ (there are too many points of $X$ to be indexed by the positive integers).  Better is to say that "$\{ N_1 ( x ) \}_{x \in X}$ is a(n open) cover of $X$."  The benefits of this will come in the second point.

*Assuming compactness you have a finite subcover, which, according to your notation should be phrased by stating that there are positive integers $i_1 , \ldots , i_k$ such that $\{ N_1 (x_{i_1} ) , \ldots , N_1 (x_{i_k} ) \}$ covers $X$.  Of course, using the notational trick in my first point we can remove one level of subscript and say instead that there are $x_1 , \ldots , x_k \in X$ such that $\{ N_1 ( x_1 ) , \ldots , N_1 ( x_k ) \}$ covers $X$.

*Also note that in your notation for the finite subcover you never mention what $m$ ranges over.

*It is completely unimportant that $N_1 (x_{k+1} )$ does not belong to $\{ N_1 ( x_{i_j} ) : j \leq m \}$; all that is important is that $x_{k+1}$ does not belong to any $N_1 ( x_{i_j} )$ for $j \leq m$.

*You appear to use $m$ as both the index that the finite subcover ranges over, and also the upper bound of the index that the finite subcover ranges over.


Sequences proof


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*I honestly have no idea what is going on with your proof.  You start with a sequence of positive reals converging to $0$, and then magically instantiate a sequence in $X$.  I'm starting to think that the underlying set of your discrete space is $\mathbb{R}$ (or something similar) which may not be the case.

*You appear to be using the same indexing of the space $X$ from the first proof.  This was very unclear to me for a while.

*I do not know what value the sequence $( 1/n )_{n =1}^\infty$ has in this proof.

*Your logical reasoning is correct (every subsequence of $(x_i)$ will have distinct terms and therefore &c.&c.&c.), however this might not satisfy someone grading a paper/exam.  Perhaps better is to say that since $d ( x_i , x_j ) = 1$ for all $i \neq j$ then no subsequence can be Cauchy (because given any subsequence $( x_{i_k} )_k$ there is no $K$ such that $d ( x_{i_k} , x_{i_\ell} ) < 1$ for all $k, \ell \geq K$) and therefore no subsequence can converge.

*The above point is contingent on "showing" that there is a sequence in $X$ consisting of pairwise distinct terms.

A: I think you're writing a much too complicated proof. Just note that $N_1(x) = \{x\}$ (which you do), so every singleton is an open set. Let $X$ be infinite and discrete. 
Supppose $X$ is compact, then the open cover $\{ \{x\}: x \in X \}$ has a finite subcover $\{x_1\},\{x_2\},\dots,\{x_k\}$. Then $X \subset \{x_1,\dots,x_k\}$, so this implies $X$ is finite, a contradiction.
As to a sequence proof, you want to show that $X$ is not sequentially compact. You write $\frac{1}{n}$, which need not be a point of $X$ (it's a general space), but the idea is there: pick a sequence $(x_n)$ in $X$ where all points are distinct, which can be done as $X$ is infinite. Suppose now that a subsequence $x_{n_k}$ converges to $x$, which means, taking $\epsilon = 1$, that for some $k \ge M$ we have $d(x_{n_k},x) < 1$, which means, by the definition of the metric that $x_{n_k} = x$ for all $k \ge M$, but this contradicts that all terms are distinct. So we have no convergent subsequence.  
