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I wrote up a proof for the continuity of y^2 for practice. Is this acceptable in the context of a Real Analysis I course?
QED
Thanks!
Thanks!
Case one is only taken care of when epsilon is big, which is the uninteresting case. The fact that delta is not a function of epsilon should be an immediate giveaway that you haven't constructed a proof
You need a < ε at the end here, but the statement looks good other than that.Let ##f:\mathbb{R} \to \mathbb{R}## be a function such that ## y \mapsto y^2##. To show continuity at every point of ##f##, it is sufficient to demonstrate:
##\forall u \;\; \forall \epsilon > 0 \;\; \exists \delta : \forall y \; \left ( |y-u| < \delta \implies |y^2 - u^2| \right ) ##
Case 1: ##u=0##
Let ##\delta = \sqrt{\epsilon}##. It is clear that ##|y| < \sqrt{\epsilon} \implies |y^2| = \epsilon##.
- ##\delta## can be a function of both ##\epsilon## and ##u## (but not ##y##) for pointwise continuity since they both proceed ##\delta##'s quantifier?
[*]Is ##(|y-u| < \delta) \wedge (\delta < |u|) \implies |y+u| < 3|u|## obvious enough to state without proof?
[*] Obviously, once ##\delta## is chosen, ##|y-u|## could be ##\geq \delta## since both were chosen arbitrarily and without respect to ##\delta##. If this inequality does not hold, neither do the following arguments. Is the parenthetical "(Note: If ##|y-u| \geq \delta## then the continuity condition is met and the argument is complete; assume ##|y-u| < \delta## holds for the remainder of the proof)" sufficient to tackle this problem?