Would a number that bounds the convergent sequence always be a bound for its limit? Given a sequence of real numbers $\left\{a_n\right\}$, that $a_n \to a$ and $a_n \leq c$. Does it follow that $a\leq c$?
I'm just perplexed by a line in the proof of the Intermediate Value Theorem in 4th Ed. of Royden.  In the last paragraph it goes like this:

Therefore $\left\{ a_n \right\} \to x_0$. By the continuity of $f$ at
  $x_0$,  $\left\{ f(a_n) \right\} \to f(x_0)$. Since $f(a_n)\leq c$ for
  all n, and the set $(-\infty, c]$ is closed, $f(x_0)\leq c$.

I just don't know where the closed $(-\infty,c]$ fits in, in the proof. 
 A: Hint: suppose $a>c$, and investigate the consequences with the $a_n$.
A: In general, any convergent sequence in a closed set will converge to a point in the set. The proof is as follows.  Let $(x_n)$ be a sequence in a closed set $A \subseteq X$ that converges to $x \in X$.  If $x \notin A$, then since $A$ is closed, $x$ is not a limit point of $A$.  Hence, there exists an open nbhd $U$ of $x$, that doesn't intersect $A$.  However, there exists $N$ such that $n \ge N$ implies $x_n \in U$, a contradiction.
Since $f(a_n) \in (-\infty, c]$, it follows that $f(x_0) \in (-\infty, c]$.
A: We proceed by contradiction. Suppose instead that $f(x_0)>c$. Then we know that $f(x_0)-\epsilon>c$ for some $\epsilon>0$. Now since $f(a_n) \to f(x_0)$, we know by the definition of convergence that there exists some $N$ such that for all $n>N$, we have $f(a_n)\in (f(x_0)-\epsilon, f(x_0)+\epsilon)$. In particular, by letting $n=N+1$, we find that:
$$
f(a_{N+1}) > f(x_0)-\epsilon>c
$$
which contradicts the fact that $f(a_n) \le c$ for all $n\in \Bbb{N}$.
A: The claim is a special case of the following fact:

Fact: If $\langle x_n:n\in\Bbb N\rangle$ is a sequence of real numbers, $\lim\limits_{n\to\infty}x_n=y$, $C$ is a closed set of real numbers, and $x_n\in C$ for all $n\in\Bbb N$, then $y\in C$.
Proof: Suppose that $y\notin C$. Then because $C$ is closed, there is an $\epsilon>0$ such that $$(y-\epsilon,y+\epsilon)\cap C=\varnothing\;.\tag{1}$$ On the other hand, $\lim\limits_{n\to\infty}x_n=y$, so there is an $m_\epsilon\in\Bbb N$ such that $|x_n-y|<\epsilon$ for each $n\ge m_\epsilon$. Pick any integer $n\ge m_\epsilon$, then $|x_n-y|<\epsilon$, so $x_n\in(y-\epsilon,y+\epsilon)$. And by hypothesis $x_n\in C$, so $x_n\in(y-\epsilon,y+\epsilon)\cap C=\varnothing$, contradicting $(1)$. Thus, $y\in C$ after all. $\dashv$

In your case $x_n=f(a_n)$, $u=f(x_0)$, and $C=(\leftarrow,c]$. To see why it’s important that $(\leftarrow,c]$ is closed, consider the following variant:

Conjecture: If $\langle x_n:n\in\Bbb N\rangle$ is a sequence of real numbers, $\lim_{n\to\infty}x_n=y$, $C\subseteq\Bbb R$, and $x_n\in C$ for all $n\in\Bbb N$, then $x\in C$.

To see that this conjecture isn’t always true, let $x_n=-\frac1{2^n}$ for each $n\in\Bbb N$, let $y=0$, and let $C=(\leftarrow,0)$, the set of negative real numbers. Certainly $x_n\in C$ for each $n\in\Bbb N$, and certainly $\lim\limits_{n\to\infty}\left(-\frac1{2^n}\right)=0$, but $0\notin C$.
The conjecture fails in this example precisely because $y$, the limit of the sequence, is a limit point of $C$ that is not itself in $C$. When $C$ is closed, this cannot happen: by definition a closed set contains all of its limit points. When $C$ is not closed, there is at least one limit point of $C$ that is not in $C$, and if such a point is the limit of the sequence, the conjecture will fail in that case, just as in the example here.
