Let (X, $\mathcal{O}$) be a topological space. Suppose that {f$_{\alpha}$(x)}$_{\alpha \in A}$ is a family of functions on $\mathbb{R}$ that are uniformly bounded and equicontinuous. Prove that the function, f(x) = $\sup\limits_{\alpha \in A}${f$_{\alpha}$(x)} is continuous.

Attempt at solution:

Let x $\in$ X and fix some $\epsilon > 0$. By definition of sup, we have that f(x) < f$_{\alpha}$(x) + $\frac{\epsilon}{2}$ for some $\alpha \in$ A. By equicontinuity, we have that for any x $\in$X, $\delta > 0$, there is a U$_{\delta}$ neighbourhood around x such that y $\in$ U$_{\delta}$ $\Rightarrow$ $|$f$_{\alpha}$(x) - f$_{\alpha}$(y)$|$ $< \frac{\epsilon}{2}$.

I'm stuck at this part. I'm not sure how to show that $|$f(x)-f(y)$|$ $< \epsilon$ from here. Also, is the $\epsilon$ in the definition of equicontinuity dependent on $\alpha$?

  • $\begingroup$ You're assuming $X$ is metrizable, what about in a general space? $\endgroup$ – msteve Sep 28 '14 at 3:23
  • $\begingroup$ I must use the open set characterization of continuity in that case? $\endgroup$ – Pi is Exactly 3 Sep 28 '14 at 3:38
  1. $f^{-1}(a,\infty)$ is open for every $a\in \mathbb{R}$. To prove this just notice that we have

$$ f^{-1}(a,\infty)=\bigcup_\alpha f^{-1}(a,\infty), $$ and these last ones are open since the $f_\alpha$ are continuous (notice that we only need the $f_\alpha$ to be continuous and uniformly bounded for this).

  1. $f^{-1}(-\infty,a)$ is open for every $a\in \mathbb{R}$. Take $x$ such that $f(x)<a$. Choose an $\varepsilon>0$ (depending on $x$) such that $f(x)< a-\varepsilon$. By definition we have that $f_\alpha(x)< a-\varepsilon$ for every $\alpha$. By equicontinuity there exists a neighborhood $U$ of $x$ such that for every $\alpha$ and every $y\in U$, we have $f_\alpha(y)<a-(\varepsilon/2)$. Taking the supremum over $\alpha$ we get $f(y)\leq a-(\varepsilon/2)$ and so $U\subset f^{-1}(-\infty,a)$.
  • $\begingroup$ Thank you. May I ask how you arrived at the equality, f$^{-1}$(a,$\infty$) = $\bigcup\limits_{\alpha}$f$^{-1}$(a,$\infty$)? $\endgroup$ – Pi is Exactly 3 Sep 28 '14 at 5:55
  • $\begingroup$ If $x\in f^{-1}(a,\infty)$ then $f(x)>a$ so that, for some $\alpha$ we must have $f_\alpha(x)>a$ by definition of $f$. This implies $x\in f^{-1}_\alpha(a,\infty)$ for this $\alpha$. For the other inclusion notice that $f(x)\geq f_\alpha(x)$ for every $\alpha$ so that $f^{-1}_\alpha(a,\infty)\subset f^{-1}(a,\infty)$. $\endgroup$ – Jose27 Sep 28 '14 at 6:02

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