# Uniformly convergence of the continuous functions sequence, $f_n$

Let $$f_n, f : D \to \mathbb{R}$$ with $$A(\neq \phi) \subset \mathbb{R}$$

Here the $$f_n$$ are the sequence of the continuous functions.

I already knew the fact "$$f_n$$ is uniformly converge to $$f$$ on $$D$$ $$\Rightarrow$$ $$f$$ is continuous on $$D$$"

But There is a lecturer's claim that

Say $$f_n$$ is continuous on $$A$$

"$$f_n$$ is uniformly converge to $$f$$ on $$A$$ $$\not\Rightarrow$$ $$f$$ is continuous on $$A$$" (statement (*))

He suggested the counter-example as like the below.

$$f_n, f : \mathbb{R} \to \mathbb{R}$$ with $$A(\neq \phi) \subset \mathbb{R}$$

Here the $$D = \mathbb{R}, A = \mathbb{Q}^c$$ and

Since the rational number set,$$\mathbb{Q}$$ is countable so $$\mathbb{Q} = \{x_1, x_2, .... \}$$

Then Defining each functions like the below

$$f_n(x) = \begin{cases}1 & \text{x \in \{ x_1, x_2, ...,x_n\}} \\ 0 & \text{o.w.}\end{cases}$$

$$f(x) = \begin{cases} 1 & \text{x \in \mathbb{Q} } \\ 0 & \text{x \in \mathbb{Q}^c} \end{cases}$$

From here, my question begins.

First question) I put in to the sequence $$x_{n+1} \in \mathbb{Q}$$ to $$\Vert f_n(x) - f(x) \Vert$$

Then $$\Vert f_n(x_{n+1}) - f(x_{n+1}) \Vert$$ = $$\Vert 0-1 \Vert = 1 \geq \epsilon(= {1 \over 2})$$

So my conclusion is this counterexample is totally wrong, since $$f_n$$ is not uniformly converge to $$f$$. What do you think about that? Is my thought wrong?

Second question) Like the lecture's thought, I guess the statement (*) is false. But I can't find any counterexample. If the lecture's counterexample is false, Please give me a counterexample.

p.s.) If the statement is true, Why does the statement(*) hold?

Thanks.

There seems to be no connection between $$D$$ and $$A$$ in $$(*)$$. So $$(*)$$ is correct. For one example take $$D=[0, \frac 12 ], A=[0,1]$$, $$f_n(x)=x^{n}, f(x)=0$$ for $$x<1$$, $$f(1)=1$$. Then $$f_n \to f$$ uniformly on $$D$$ and $$f$$ is not continuous on $$A$$.
However, the example given by the lecturer is wrong since $$f_n$$ does not converge uniformly to $$f$$ on $$D=\mathbb R$$.
Answer for the revised question: If $$f_n \to f$$ uniformly on $$A$$ and each $$f_n$$ is continuous on $$A$$ then the restriction of $$f$$ to $$A$$ is continuous. But you cannot say that $$f$$ is continuous at points of $$A$$ as in the example by your lecturer.
• Thanks for reply, Dear Mr,@Murthy. But I should have been asked "$f_n$ is uniformly converge to $f$ on $A$" in statement (*) not "$f_n$ is uniformly converge to $f$ on $D$" Sorry for typo. I edited my question. – se-hyuck yang Jan 28 '20 at 8:25
• What about the case if we add the condition $A \subset D$? Though adding more condition, his claim and counterexample are still not hold, right? – se-hyuck yang Jan 28 '20 at 14:31
• @se-hyuckyang Actually $f_n(x)=f(x)=0$ for all $x$ in $A$ in his example. So $f_n \to f$ uniformly on $A$. – Kavi Rama Murthy Jan 28 '20 at 23:15