2
$\begingroup$

Let $(X, d)$ be a compact metric space.

Suppose $f$ and $(f_n)$ are real-valued continuous functions on $X$.

Suppose that, for each $x\in X$, the sequence $(f_n(x))$ is a monotonic sequence converging to $f(x)$.

Show that $(f_n)$ converges uniformly to $f$.

(Hint: Given $\epsilon > 0$, show that the sets $U_n = \{x \in X : | f_n(x) − f (x)| < \epsilon\}, n \in \mathbb{N}$, form an open cover of $X$)

I do not really understand what this hint can do to solve the problem. Thank you.

  • 1
    Hint : continuous +compact =uniformity continuous – 2017-01-09
  • 0
    @vvnitram what I know about uniformly continuous for now is the definition. – 2017-01-09
  • 1
    @R.T., the hint says to form a countable open cover. This is so that you can next use the compactness of $X$ to extract from that cover a finite subcover, $U_{n_{1}}, U_{n_{2}}, \ldots, U_{n_{K}}$. So, for example, $|f_{n_{1}} - f| < \epsilon$ on all of $U_{n_{1}}$. But then, it follows (by the monotonicity that you are given) that $|f_{m} - f| < \epsilon$ on all of $U_{n_{1}}$ for all $m > n_{1}$. – 2017-01-09

2 Answers 2

2

Let $\epsilon > 0$.

Once you have $U_n = \{x \in X : | f_n(x) − f (x)| < \epsilon\}, n \in \mathbb{N}$, form an open cover of $X$, since $X$ is compact, there is an $N \in \mathbb{N}$ such that $X \subset \bigcup_{i=1}^{N}U_i$. Since $U_i \subset U_j$ for $i \leq j$, then $\bigcup_{i=1}^{N}U_i=U_N$, and hence $X \subset U_N \subset U_{N+1} \cdots$

Therefore, for all $x \in X$ we have $|f_n(x)-f(x)|< \epsilon$ for all $n \geq N.$

  • 0
    So it has nothing to do with $(fn(x))$ is a monotonic sequence converging to $f(x)$? – 2017-01-09
  • 1
    That guarantees $U_i \subset U_j$ for $i \leq j$. – 2017-01-09
  • 0
    Can you give me a brief explanation of how monotonic guarantees $U_i\subset U_j$? Thanks. – 2017-01-09
  • 0
    If $|f_N(x)-f(x)| < \epsilon$, since $f_n$ is monotonic and converges to $f$, then $f_{N+1}(x)$ can't be further away from $f(x)$, that is, $|f_{N+1}(x)-f(x)| \leq \epsilon$, – 2017-01-09
1

Well, since $X$ is compact you only need a finit enumber of elements to cover the whole set. Since $U_n \subseteq U_m$ for $n\leq m$ we have that there is an $N$ such that $U_N,U_{n+1},\dots $ are all equal to $X$, which is what you wanted to prove.

  • 0
    I do not really follow your logic. Since $X$ is compact, $\exists U_n'\subseteq U_n$ finite s.t. $X\subseteq U_n'$. That's what I get. Can you explain what do you mean staring from $U_n\subseteq U_m$? – 2017-01-09
  • 1
    @R.T. That's not what compact means. It means that if you have a (possibly infinite) collection of open sets that together cover $X$, there is a finite subcollection that also covers $X$. The sets themselves are rarely finite. Since, by definition of convergence, every $x\in X$ is in _some_ $U_n$, the infinite collection of $U_n$ covers $X$. By compactness, it is possible to pick a finite number of $U_n$'s that cover $X$. Since every $U_n$ is contained in the ones that come after, it suffices to pick, from this finite collection, the largest one, and it will cover $X$. – 2017-01-09
  • 0
    @Arthur Ok I see. Thanks. – 2017-01-09