removing a contact

[iso300]

I was wondering if I'm able to remove my contact on another person's MSN
Messenger list.

Until today I realized that a particular individual who I have added to my
list on two separate MSN Messenger accounts has disappeared from my list. I
never deleted/removed any contacts. Even if he has removed me from their
list, I should still be able to see that individual on my list right
(perhaps not in Online mode)? When I will right-click on their name, the
Delete option will then become available correct?

Just to confirm, I'm sure that I didn't delete that individual's contact
from my lists but his name does not appear on my list anymore. Plus, when I
went to see the list of people who added me, his name is not on it either.

Thanks for your replies.

[Jonathan Kay [MVP]]

Greetings,

No, you can't really affect anyone's contact list -- nor can they "remove" themselves from
yours. If you open up the main Messenger window, click the Tools menu, then Options, then
Privacy tab, can you find his name in there?
____________________________________________
Jonathan Kay
Windows MVP, Messenger
Associate Expert
http://www.microsoft.com/windowsxp/expertzone/
Messenger Resources - http://messenger.jonathankay.com


"iso300" > wrote in message
ble.rogers.com...
> I was wondering if I'm able to remove my contact on another person's MSN
> Messenger list.
>
> Until today I realized that a particular individual who I have added to my
> list on two separate MSN Messenger accounts has disappeared from my list. I
> never deleted/removed any contacts. Even if he has removed me from their
> list, I should still be able to see that individual on my list right
> (perhaps not in Online mode)? When I will right-click on their name, the
> Delete option will then become available correct?
>
> Just to confirm, I'm sure that I didn't delete that individual's contact
> from my lists but his name does not appear on my list anymore. Plus, when I
> went to see the list of people who added me, his name is not on it either.
>
> Thanks for your replies.
>
>
>