# Open and closed sets

## Homework Statement

For the following set state and justify:

i) whether or not it is open?

ii) whether or not it is closed?

{z||z|<2}

## The Attempt at a Solution

I really don't know where to start on this question, i don't understand this topic at all and any help starting would be great.

thanks

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Hootenanny
Staff Emeritus
Gold Member
Do you know what open and closed sets are?

well a set is closed if its complement is open.

i don't really know what open and closed is though.

Hootenanny
Staff Emeritus
Gold Member
well a set is closed if its complement is open.

i don't really know what open and closed is though.
That's correct. Roughly speaking a set is closed if the set contains the boundary of that set.

For example, consider a set defining a sphere. If the set includes all points inside the sphere, as well as the surface, then the set is closed. However, if the set only contains point on the interior (i.e. inside) the sphere, but excludes the surface, the set is closed.

Similarly for the real axis, {x:0<x<1} is an open set since you can have all numbers between zero and one, but not including zero and one themselves. Conversely, {x: 0 ≤ x ≤ 1} is closed since x can take any value between zero and one, including zero and one themselves.

Do you follow?

A set Y is open if every point in Y is an interior point
A set Y is closed if every point in Y is an limit point

yes....i understand all of that. i don't really understand what the set {z||z|<2} actually means though.

Hootenanny
Staff Emeritus
Gold Member
yes....i understand all of that. i don't really understand what the set {z||z|<2} actually means though.
In words it means the set of all z such that the magnitude of z is less than one. Geometrically, what does this set look like in the complex plane?

did you mean to say the magnitude of z is less than two?

then in the complex plane it would be a circle? with radius two? or would it have to be less than two?

Hootenanny
Staff Emeritus
Gold Member
did you mean to say the magnitude of z is less than two?
Yes I did, thanks.
then in the complex plane it would be a circle? with radius two? or would it have to be less than two?
You're on the right lines, the set contains all points in the complex plane who's magnitude is less than two. So yes, this does look like a circle of radius two, but the question is: does the set include the boundary (i.e. the points on the circumference) of the circle?

the set wouldn't include the boundary, due to |z|<2, which means that the set is open.

Hootenanny
Staff Emeritus
Sounds good to me 