Question

\[(M,d), (N,r) \text { metric spaces } f: M\rightarrow N \text{ continuous at } x_0, f(M)\subset B\subset M\] is
\[f\text{ continuous as a function } (Md) \text {to} (B,r)\] \[(M,d), (N,r) \text { metric spaces } f: M\rightarrow N \text{ continuous at } x_0, f(M)\subset B\subset M\] is
\[f\text{ continuous as a function } (Md) \text {to} (B,r)\] @Mathematics

Answer 1

is this in carothers too?

Answer 2

Did you typed the question correctly.

Answer 3

no it is not from corothers, and i think i typed it in right.

Answer 4

I find this part a little weird

\[f(M)\subset B\subset M\]

Answer 5

ok that is a typo too. should be
\[f(M)= B \subset N\]

Answer 6

in other words, is the function continuous if you restrict to the codmain. my initial reaction is "of course" but frequently there are pathologies i don't take in to account. but if the pre-image of an open set is open, i can't see how the restriction can make any difference.

Answer 7

I would try just using the epsilon delta definition of continuity on a metric space.. Of hand I can see any problems either

Answer 8

off

Answer 9

are you just checking for continuity at x0 correct?

Answer 10

yeah that is what i am thinking. sometime the obvious thing is in fact true. i will look at this again when the wine wears off. thanks again.
yes and x0

Answer 11

i was worried about what the different metrics could be, but open mean open irrespective ov the metric, so i cannot see a problem

Answer 12

you know what I meant...but I meant to type...can't see any problems

Answer 13

especially since you are not changing the metrics themselves...just the underlying set

Answer 14

right. so i am going to stick to the obvious answer, namely that the preimage of and open set is open, and the image is in the codomain.

Answer 15

actually that can change which sets are open, but I don't see that as a problem here

Answer 16

but topological definition is pre-image of open set is open. pre-image live in F(M) so it should be fine.

Answer 17

with respect to (B,r) B is clopen

but B might not be open in N

Answer 18

yes but the premise is that f is continuous at x0 as a function form M to N

Answer 19

correct...so I don't think it matters.

Answer 20

oh i see what you are saying. maybe B is not open it N? but it must be right?

Answer 21

no

Answer 22

inverse image of a neighbor of f(x0) is open in N .

Answer 23

but is it possible it is not open in f(M)
?

Answer 24

it will be

Answer 25

ok that is what i thought. thanks again. now it is dinner time.

Answer 26

Thanks for posting these problems. I don't get many opportunities to talk to someone about real analysis.