MHB Proving Uni.Conv. of Fn to F on (0,1): Questions & Answers

[bw0young0math]

Hello, today I have a question.

If uni.cont. function sequence fn on (0,1) is uni.conv. to f on (0,1), then f is uni.cont. on (0,1).

The above is true.
The wonder I have is...May I prove the above by this way?

This way: fn is uniform continuous on (0,1). So Fn is continuous on [0,1] (by continuous extention theorem.)
If I can prove Fn is uni. convergent to F on [0,1], (since Fn is cont. on [0,1]) F is cont. on [0,1] and f is uni. cont. on (0,1).

is this way true?
IF it is true, how can I prove Fn is uni.conv. to F on [0,1]?
IF not, could you show me some counter example?Please...(Sadface)(Sadface)(Sadface)

 

[Ackbach]

I think you can use a standard $\varepsilon/3$ argument to prove this. Write out what uniform continuity and uniform convergence mean, using $\epsilon/3$ as the RHS. Then use the Triangle Inequality to finish.