Proving Convergence of Absolute Value Sequence in Real Numbers

[Daveyboy]

For a sequence in the reals

{a_n} converges to a, show {|a_n|} converges to |a|.

For any e>0 the exists an N s.t. for any n>N |a_n-a|<e

I want to use this inequality, but there is something funny going on. I do not know how to justify it.

|a_n-a|\leq||a_n|-|a||

 

[HallsofIvy]

Look at three separate cases.

1) a> 0. Can you show that, for some N, for all n> N a_n&gt; 0? (Take \epsilon= a/2.)

2) a< 0. Can you show that, for some N, for all n> N a_n&lt; 0?

3) a= 0. Here, ||a_n|- a|= ||a_n||= |a_n|.

 

[Daveyboy]

Okay I see how to break it down case wise and find N accordingly. That will work nicely.

However, I was hoping to use the reverse triangle inequality but I run into the double abs. value. It just doesn't look right to say that
for any e>0 there exists and N s.t. for any n >N

|a_n - a| < e
and
|a_n - a| \geq |a_n| - |a|
implies

e>||a_n| - |a||

but if I showed this wouldn't it be true?