# I understand deltas and epsilon proofs for the most part

1. Mar 21, 2010

### Reptar

so 0 < l x-a l < delta and l f(x)-L l < epsilon

What I don't understand is how come deltas and epsilons can't be greater than or equal to their respective differences?

2. Mar 21, 2010

### mrbohn1

Convention? One less line to draw!

3. Mar 21, 2010

### HallsofIvy

Because it is "less than" that is important- and if I can find $\delta$ so that $0< |x- a|\le \delta$ I could always choose $\delta$ just slightly larger and have $0< |x- a|<\delta$.

4. Mar 21, 2010

### Tinyboss

Some writers (though not many) even follow the convention that < means "less than or equal to", and reserve $$\lneq$$ for strict inequality. More commonly, $$\subset,\subseteq$$ are used interchangeably for set inclusion, with proper inclusion indicated by $$\subsetneq$$.

It's not a bad convention IMO, because, for instance, for a sequence converging to L, you can write $$\forall n,a_n<c\Rightarrow L<c$$, without worrying that limits don't preserve strict inequality.

5. Mar 22, 2010

### snipez90

Dang why is that a good convention? Maybe that just looks weird to me. Anyways I'm perfectly content with just knowing that when I pass off to limits, then I need the non-strict inequality sign. These are rather nitpicky things that I cared way too much about when I started learning analysis. Also, I don't understand the point of getting a nice "less than epsilon" end to an argument, though I admit sometimes it's maybe worth the few extra minutes to finish with "< e" instead of "< e(some ugly factor)".