Let $a_n$ be a convergent sequence wit limit $L$. With out using the Heine-Borel theorem, prove that the set $\{L,a_1,a_2,...\}$ is compact. Let $a_n$ be a convergent sequence wit limit $L$. With out using the Heine-Borel theorem, prove that the set $\{L,a_1,a_2,...\}$ is compact.
I know that $a_n$ is convergent to $L$, meaning for all $\epsilon >0$, there exists an $N>0$ such that 
$n>N$   implies $|a_n -L|< \epsilon $
I also know that $a_n$ is convergent, so it's bounded, so is $\{L,a_1,a_2,...\}$ . I can prove this set is both bounded and closed, so it's sequentially compact, hence compact. But I'm not allowed to use this theorem. How can I use the above info to prove that set is compact?
 A: Given the set $\{L,a_1,a_2,...\}$ take a covering by open sets  $\{U_i, i \in I\}$. 
One of the $U$ namely $U_{i_L}$ contains the point $L$. Being $L$ a point of $U_{i_L}$, because $U_{i_L}$ is open exists $r>0$ which $B(L,r) \subset U_{i_L}$. 
In the convergence definition for $a_n$ take $\epsilon=r$, then exists $N \in \mathbb{N}$ which $ \forall n\geq N$ implies $|a_n-L|<r$. That means every point of the succesion for $n\geq N$ is contained in the ball (because the distance to the center $L$ is less than $r$), i.e.: $$a_n \in B(L,r), \forall n \geq N$$
and therefore in the set $U_{i_L}$.
Each point in the succesion with $n<N$ is contained in at least one of the $\{U_i, i \in I\}$, for each one of they pick one $U_{i_n} \in \{U_i, i \in I\}$ with the property $a_n \in U_{i_n}$. Then $a_1,...a_{N-1}$ is covered by $\{U_{i_1},U_{i_2},...,U_{i_{N-1}}\}$.
Finally the covering $\{U_{i_1},U_{i_2},...,U_{i_{N-1}},U_L\}$ is a finite covering of the set $\{L,a_1,a_2,...\}$ because the last one covers the succesion for $n\geq N$ (and the limit point $L$) and the first ones covers the beginning of the succesion.
A: Suggestion: For each $ε>0$ there is $N \in \mathbb N$ so that $$a_n \in B_ε(L)$$ for all $n\ge N$. In words, there are only finite many $a_k$ that are not in this open ball around $L$. Now take the open cover $$\mathcal O=\left(\bigcup_{k=1}^{N-1}B_{δ_k}(a_k)\right)\cup B_ε(L)$$ which contains $\{L,a_1,a_2,\ldots\}$, in symbols $$\{L,a_1,a_2,\ldots\} \subset \mathcal O$$This is finite and minimal in the sense that you can choose the constants $δ_k$ and $ε$ arbitrarily small. So every open cover $\mathcal U$ of $\{L,a_1,a_2,\ldots\}$ contains the finite subcover $\mathcal O$. 
(Although, this last sentence "So..." can require a better justification. But if $\mathcal U$ is a random open cover then there are open balls around each $a_n$ and $L$ that contain them. So this balls contain the balls defined in $\mathcal O$ and the conclusion follows).
Otherwise, if you were allowed to use that compact means closed and bounded, you could note that every subsequence $\left(a_{n_j}\right)_{j\in J}$ of the sequence $\left(a_n\right)_{n\in \mathbb N}$ converges also to $L$ (subsequence of convergent sequence), which makes $L$ the single limit point of the given set.  
A: A set $A$ is compact in a space $X$ if given any arbitrary collection of open sets $\{G_{\lambda}\}$ such that $A \subseteq \bigcup G_{\lambda}$ then there is a finite collection of sets from $\{G_{\lambda}\}$, say $\{G_1, G_2, ..., G_n\}$ such that $A \subseteq \bigcup_{i = 1}^n G_i$. What this essentially means is if there is say a infinite collection of open sets which contain $A$ then there are a finite number of sets from the said collection which also contain $A$. Example, Say $x \not \in A$. Then the collection of sets $\{B_n \ | \ n \in \Bbb N\}$ where $B_n = \{y \ | \ || y - x|| \lt n\}$ where contains $A$. If $A$ is compact there are finite number of sets $B_{m_1}, B_{m_2}, \cdots B_{m_n}$ such that $A$ is contained in the union $\bigcup_{i = 1}^n B_{m_i} $. This is the primary definition of compactness. The "bounded and closed" definition is the Heine-Borel Theorem. Read more here. A collection of sets of the form $\{G_{\lambda}\}$ is called an open cover of $A$. The reason we use the subscript $ \lambda$ here is to emphasise that the collection $\{G_{\lambda}\} $ can be infinite and uncountable in fact. 
Now to prove that a set $A$ is compact we must prove every open cover of $A$ contains a finite subcover. The terminology here should be clear due to the above example. So we must assume an arbitrary open cover of $A$ and prove there are a finite number of sets from that collection (the open cover) which also contain $A$. Let us get to it. 
Let $A = \{L,a_1,a_2,...\}$. Assume there is an open cover $\{G_{\lambda}\}$ - as explained above -  such that $A \subseteq \bigcup G_{\lambda}$. Since $L \in A$ there is a set $G_{\alpha} \in \{G_{\lambda}\}$ such that $L \in G_{\alpha}$. Now by definition $G_{\alpha}$ is an open set. By definition a set $Q$ is open if for each $x \in Q$ there is $\epsilon \gt 0$ such that $ \{y \ | \ |y - x| \lt \epsilon \} \subseteq Q $. This follows from the fact that every open set (in $\Bbb R$ if you will) contains a neighbourhood of each of its points, by definition. So since $L \in G_{\alpha} $there is an $\epsilon \gt 0$ such that $  \{y \ | \ |y - L| \lt \epsilon \} \subseteq  G_{\alpha}$. Now since $L$ is the limit of the sequence $(a_n)$ given  any $\epsilon \gt 0$ there is an $m \in \Bbb N$ such that $ n \ge m \implies | a_m - L | \lt \epsilon$. Therefore we can infer that all elements in $(a_n)$ whose index is larger than $m$ are in $ \{y \ | \ |y - L| \lt \epsilon \} $. Therefore we get that 
$$ \{L, a_m, a_{m + 1}, a_{m + 2}, a_{m + 3}, \cdots\} \subseteq \{y \ | \ |y - L| \lt \epsilon \} \subseteq  G_{\alpha} ----- (1) $$
Now we are almost done. We have shown that every element $a_n$ for $n \ge m$ is contained in exactly one specific set - the one which also contains $L$ - from the collection $\{G_{\lambda}\}$. Now all we need is to prove that the rest of the elements in $A$ are also contained in a finite number of sets in the collection $\{G_{\lambda}\} $. This is easy since $a_i \in A \implies a_i \in \bigcup G_{\lambda} \implies a_i \in G_i $ where $G_i \in \{G_{\lambda}\}$ is some set in the collection. And this is true for each $i = 1 ... m - 1$. So by $(1)$ the collection $\{ G_{\alpha}, G_1, G_2, ...G_{m - 1} \}$ forms a finite sub-collection of sets from $\{G_{\lambda}\}$ whose union contains $A$. 
$\{G_{\lambda}\}$ here was chosen arbitrarily so this is true for any open cover of $A$ and hence by definition $A$ is compact. 
Hope this helped.   
A: Let $\{U_i\}_{i\in I}$ be an open cover of the set $\{L,a_1,a_2,\dots\}$. There exists some index $i\in I$ such that $U_i$ contains $L$. The fact that $a_n\to L$ implies that $U_i$ will contain all but finitely many of the $a_n$, or in other words, $U_i$ will contain $a_n$ for $n\geq M$ for some $M\in \mathbb N$. That's by the definition of convegence in topological spaces ( see http://en.wikipedia.org/wiki/Limit_of_a_sequence). 
Now, take the sets $U_{j_n}$ that contain some $a_n$ for $n=1,\dots,M-1$, and you have covered everything using
$$U_i \cup \bigcup_{n=1}^{M-1}U_{j_n}$$
