Topology #11 Metric Equivalences

Equivalence of two Metric Topologies

[ba]

Thank you so much for awesome video. This is really helpful in concise and clarity to subject matter. I must say these videos were so addictive in intuitively made that learning was fascinating. Appreciated!

[valyanurfadila]

I just found somewhere else that two metrics are said to be equivalent if there exists a, b > 0 so that
a * d1(x, y) <= d2(x, y) <= b * d1(x, y) .

Is that definition true though ?

I wonder if there's anyone kindly enough to help me with this example of partial metric space (nonzero self-distance) that I'm working on :

Given a partial metric space (X, p) and d1, d2 : X × X→ R+, where

d1(x, y) = 2p(x, y) − p(x, x) − p(y, y), ∀x, y∈X and
d2(x, y) = max{ p(x, y) − p(x, x), p(x, y) − p(y, y) }, ∀x, y∈X.

note : (p is a partial metric)

How do I prove that d1(x, y) and d2(x, y) are equivalent?

[vaidyt]

I couldn't understand one thing:
in the end I thought you need to prove that:
d2(x, y) <= d1(x, y)*k2 (because 2 is max metric and 1 is euclidean metric).
but what you have proved is the reverse... am confused.. plz help..

[PLUSSLife]

@vaidyt this is what is confusing me, it seems like he keeps on proving that the max is greater than the euclidean..

[vincent.p]

i would have said
d1 ~ d2  then Tau1 = Tau2
am i wrong ?

[laflaca]

ok, this is the topological equivalence of metrics, not the strong one, i mention this because that conffused me a lot

[aranbercu]

I certainly understand what you are saying here, but there are simpler ways to explain it.