Right. But you could have been a bit more explicit here. For example you could have argued like this:
Originally Posted by arbolis
But note: by arguing like this you already assume that f restricted to A happens to be continuous, so that there is no reason to even try to prove it only for points on the line the way you do below:
A is the half-plane below (and including) the line
. f happens to be continuous at any inner
point of either A or
: because f happens to be identical to a continuous function in an entire neighborhood of such a point.
Thus, the only problematic points are those on the line
Maybe you find it difficult to prove that goes to as tends to . Initially I wouldn't have thought of even trying to prove this: one usually assumes that a polynomial function of the coordinates of a point of is continuous without further addo. This is because the projection of a point of onto its k-th coordinate is continuous, and that sums and products of continous functions are continuous.
I'll try to prove it is :
I want to prove that
such that if
which work if I take
. Hmm no it doesn't work. I must sleep now, will think about it.
If you have any help to provide, it's always a pleasure for me. Don't hesitate.
Thus there is no reason to prove continuity of a polynomial in the coordinates for each and every such polynomial using a pure delta-epsilon argument: it would be too much work (depending on the complexity of the polynomial) and it wouldn't be particularly enlightening either.