Is an accumulation point of a set $A \subset C^*(X)$ always continuous? Definitions: Let $C(X)$ be the space of continuous real valued functions of $X$, $C_p(X)$ be the space of real valued functions of $X$ with the topology of pointwise convergence. Denote $C^*(X) = \{ f \in C(X) | f \ is \ bounded \}$. Let $C_p^*(X)$ be the set $C^*(X)$, with the topology inherited from $C_p(X)$.
We also say that $x$ is an accumulation point of $A$ in $X$, if for every open neighborhood $U$ of $x$, we have $U \cap A \neq \emptyset$.
My question is:
Suppose that $f \in \mathbb R^X$ is an accumulation point of $A \subset C^*(X)$. Can we conclude that $f$ is continuous? and, can we conclude that $f$ is bounded?
Thank you!
 A: A limit of a sequence is the accumulation point of the set of terms of the sequence. The most known counterexample to show that $C(X)$ is not complete is $f_n(x)=x^n$, $x=[0,1]$ which is pointwisely convergent to $f(x)=\lfloor x\rfloor$, that is not continuous.
A: It suffices to notice that convergence of a sequence/net in $\mathbb R^X$ with the product topology is precisely the pointwise convergence.
Therefore any non-continuous function, which is a pointwise limit of a sequence of bounded continuous functions, shows that continuity is not preserved. You can take, for example
$$f_n(x)=
  \begin{cases}
    x^n & x\in[-1,1], \\
    1 & x\ge1, \\
   -1 & x\le-1.
  \end{cases}
$$
You also ask about boundedness. If we take
$$f_n(x)=
\begin{cases}
  x & x\in[-n,n],\\
  n & x\ge n,\\
 -n & x\le -n
\end{cases}$$
then this is a sequence of bounded continuous functions, where the limit is not bounded.
(You could combine these two examples to get a sequence of bounded continuous function such that the limit is both unbounded and non-continuous.)

I'll just mention that if you take the set of all functions which are bounded by some fixed constant $M>0$, i.e.
$$A_M=\{f\in \mathbb R^X; |f(x)|\le M\}$$
then any accumulation point of $A_M$ belongs to $M$. (I.e., when you ask about the boundedness, it would work if all functions would be bounded by the same constant.)
If a point is an accumulation point of some set, then there exists a net consisting of points of this set, which converges to the said point.
In our case, every accumulation point of $A_M$ must be a limit of a net $(f_\lambda)_{\lambda\in\Lambda}$, where $|f_\lambda(x)|\le M$ for each $x$ and each $\lambda$. If this is true, then also $|f(x)|\le M$, where $f=\lim f_\lambda$ is limit of this net. 
Since $A_M=[-M,M]^X$, we basically just repeated one of possible arguments in the proof that product of closed sets is closed.
