Because @isec, @diff, @union and @aonly will not be effected by that assignment but you seem to think that they will.

I just put this code through.

my @x = ("1");
my @y = ("1");
my @z = ("1");
print "Before: @x, @y, @z\n";
(@x, @y, @z) = ();
print "After: @x, @y, @z\n";
[download]

and got this output:

Before: 1, 1, 1
After: , ,
[download]

It initialized all variables.

It initialized all variables.

Yes, but that's completely different than what people are trying to tell you. If you write:

my @x;

... @x contains the same thing as if you had written:

my @x = ();

... because @x will contain no elements. People are trying to tell you not to waste time and keystrokes with the latter, because it's unnecessary and a distraction.