Show that $\lbrace x_n : n \in \mathbb{N} \rbrace \cup \lbrace x \rbrace$ is a compact subset of $(X,d)$

Question

Let $(X,d)$ be a metric space, and $(x_n)_{n \in \mathbb{N}}$ a sequence in $X$, with $x_n \rightarrow x \in X$ (w.r.t the usual metric $d$). Show that $\lbrace x_n : n \in \mathbb{N} \rbrace \cup \lbrace x \rbrace$ is a compact subset of $(X,d)$

Need step by step proof!

These are my thoughts:
I know to show $(X,d)$ is a compact subspace I have to show that every open cover of $X$ has a finite subcover.

This means I must first show $(X,d)$ is an open cover of (what?)
But, isn't the singleton $\lbrace x \rbrace$ closed in $(X,d)$? Also, I suspect this a gross misunderstanding, but if $x_n \rightarrow x \in X$ (w.r.t the usual metric $d$), then is it wrong to interpret the union in question to be:

$\lbrace x \rbrace \cup \lbrace x \rbrace$?

Answer 1

For an open cover of $\{x_n\mid n\in\mathbb{N}\}\cup\{x\}$ there is an open set $U$ from the cover with $x\in U$ and because $x_n\to x$ there is a $N\in \mathbb{N}$ with $x_m\in U$ forall $m>N$. So when we chose for every $m\leq N$ an open set $U_m$ from the cover with $x_m\in U_m$ we get a finite subcover by the open sets $U_m, m\leq N$ and $U$ and so $\{x_n\mid n\in\mathbb{N}\}\cup\{x\}$ is compact.

Answer 2

Hint: Let $S:=\{x_n: n\in \mathbb N\}\cup \{x\}$. Since $(X,d)$ is a metric space, then showing that $S$ is compact is equivalent to showing that $S$ is sequentially compact: that every sequence in $S$ has a convergent subsequence whose limit is in $S$. Now the convergence of $x_n$ to $x$ implies that any subsequence of $x_n$ converges also to $x$.

Answer 3

If $\lim_{n\to \infty}d(x,x_n)=0,$ then, by the def'n of $\lim$, whenever $r>0$, the set $\{n: x_n\not \in B_d(x,r)\}$ is finite (where $B_d(x,r)=\{y: d(x,y)<r\}$).

$U$ is an open subset of $X$ iff for every $y\in U$ there exists $r_y>0$ such that $B_d(y,r_y)\subset U.$

For brevity let $A=\{x\}\cup \{x_n: n\in N\}.$ Let $C$ be an open cover of $A.$ There exists $U_0\in C$ with $x\in U_0,$ and there exists $r_x>0$ such that $B_d(x,r_x)\subset U_0$.

Let $S=\{n:x_n\not \in B_d(x,r_x)\}.$ Then $S$ is finite. For each $n\in S$ there exists $U_n\in C$ such that $x_n\in U_n.$

Then $C^*=\{U_0\}\cup \{U_n:n\in S\}$ is a finite subset of $C.$ And $C^*$ is a cover of $A .$ Because: (i). $x\in U_0.$ (ii) If $n\not\in S$ then $x_n\in B_d(x,r_x)\subset U_0.$ (iii). If $n\in S$ then $x_n\in U_n.$

Remark. It is useful that, in a metric space, a sequence $(x_n)_{n\in N}$ converges to $x$ iff $\{n:x_n\not \in U\}$ is finite for every nbhd $U$ of $x.$

Answer 4

In a metric space $X$, let $x_n$ be a sequence that converges to the element $a$. Then the set $M=\left\{a, x_1, x_2, \ldots\right\}$ is compact.

Solution 1.

We consider that the set $M$ is cover compact (compact). Let $(O_\lambda)_{\lambda \in \lambda}$ be an open covering of $M$, that is, every set $O_\lambda \subseteq X$ is open and, moreover, it holds that $$ M \subseteq \bigcup_{\lambda \in \lambda} O_\lambda. \tag{1} $$ To prove that $M$ is cover compact, we need to show that $M$ can only be covered by finitely many of the open sets (finite partial cover). First, because of (1), there is an index $\lambda_0 \in \lambda$ with $a \in O_{\lambda_0}$. Thus $a$ is an interior point of the open set $O_{\lambda_0}$, which is why we find a number $\varepsilon>0$ with $B_{\varepsilon}(a) \subseteq O_{\lambda_0}$. As usual, here $B_{\varepsilon}(a)$ denotes the open ball around $a$ with radius $\varepsilon$.

Since the sequence $\left(x_n\right)_n$ converges to $a$ by premise there is by definition a natural number $N \in \mathbb{N}$ which depends on $\varepsilon$, with $$ d(x_n,a)<\varepsilon \quad \text{ or equivalently } \quad x_n \in B_{\varepsilon}(a) \quad \forall\, n \geq N. $$ But this means just $x_n \in O_{\lambda_0}$ for all $n \geq N$. Thus, except for finitely many sequence members, all members lie in the open set $O_{\lambda_0}$. Finally, because of (1), for every $j \in\{1, \ldots, N-1\}$ we find an index $\lambda_j \in \Lambda$ and an open set $O_{\lambda_j}$ with $a_j \in O_{\lambda_j}$. Let us now write $$ M=\{x_1, \ldots, x_{N-1}\} \cup\{x_n \in X \mid n \in \mathbb{N} \text { with } n \geq N\} \cup\{a\}, $$ so with our considerations it follows as desired $$ M \subseteq \bigcup_{j=1}^{N-1} O_{\lambda_j} \cup O_{\lambda_0}, $$ that is, we have found a finite partial cover of the set $M$.

Thus, as desired, we have shown that the set $M$ is compact.

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Solution 2.

We will prove that the set $M$ is sequence compact (compact). To do this, we need to show that every sequence from $M$ has a convergent subsequence. To this end, let $(y_n)_n \subseteq M$ be any sequence. We distinguish the following two cases:

(a) The sequence takes a value in $M$ any number of times. Thus the sequence $(y_n)_n$ automatically has a constant subsequence which is obviously convergent.

(b) The sequence takes any value from $M$ only finitely often. Let us define for $k \in \mathbb{N}$ the set $$ M_k=\left\{x_j \in X \mid j \in\{1, \ldots, k\}\right\}, $$ then for each $k \in \mathbb{N}$ there exists at least one index $$ N_k \in \mathbb{N}: x_n \notin M_{N_k}\quad \forall\, n \geq N_k. $$ Now let $\varepsilon>0$ be arbitrary. Since the sequence $ (y_n )_n$ converges to $a$ by precondition, $$ \exists\, K \in \mathbb{N}: d(y_n-a)<\varepsilon\quad \forall\,n \geq K. $$ Thus, for $n \geq N_K$ it follows. $$ d(y_n-a)=d(x_n-a)<\varepsilon, $$ that is, the subsequence $(y_n)_n$ also converges to $a$.

Thus, as desired, we have shown that the set $M$ is compact.