# Epsilon approximation some intuition

I am currently making my analysis knowledge better so while reading apostol I encountered the following theorem below, so my question is I want to see if my intuition is correct as always with analysis when they play with epsilon sometime you just understand the technicalities but not the actual intuition of why it works. So the theorem that they invoke in their proof that is the following statement

property specifically the following statement

Let S be a non-empty set of real numbers with a super mum, say b = sup S. Then for every a < b there is some x such that a < x $\leq$ b.

So when they show the other inequality is satisfied to get an equality they use an episolon argument, so I was thinking why not can we use the theorem above directly ? but the reason why can't use it directly is that we need to choose an element a but that element has to be in our set and so we we get our sup and minus that by arbitrarily $\epsilon$ in order to get back in our set and that is why it works is my understanding correct?

• I would love to have a super mum! Jun 1, 2015 at 22:45
• sure we would all like that too :P
– user111750
Jun 1, 2015 at 22:51

$$0<y-a$$ Remember, y-a will vary depending on what a we choose. So, the difference can vary depending on what y we choose. Now, we proceed by letting epsilon=y-a to get the folowing statement. $$a=y-\epsilon<y<b$$ Hidden in the middle there, is the most important part. For any y, there is a sufficient epsilon such that the difference between y and epsilon is less than the supremum. Hope this helps. I didn't really answer your exact question but I hope I demystified the epsilon definition of a supremum for you and help you better understand why the two statements are equivalent.