3

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!

Community
  • 1
topsi
  • 4,222

2 Answers2

2

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.

ajotatxe
  • 65,084
2

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.

Martin Sleziak
  • 53,687