A Topology - Boundary of a ball without a point

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Let us say we have f analytic in Ball_1(0).
which means, radius 1, starting at z_0 = 0 point.
If I want to find the boundary of Ball_1(0).
Will the boundary be {0} or {empty}?
Not homework, just an intuition to understand f(z)=1/z function ( for example ) better.
Let us say we have f analytic in ##Ball_1(0)##. which means, radius 1, starting at ##z_0 = 0## point. If I want to find the boundary of ##Ball_1(0)##. Will the boundary be ##{0}## or ##{\emptyset}##? Not homework, just an intuition to understand ##f(z)=\frac 1 z## function ( for example ) better.
 
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As stated, assuming real space of n dimensions, the boundary consists of all points at a distance 1 from the origin. Your statement is confusing? What has f to do with the ball boundary?
 
Hi, it does not matter anymore.
Asked someone from my course which I trust and he said to me the answer :)
Basically, As I said.
At complex analysis.
If you have the Ball I said, with radius 1 and it beginning at point zero.
If you create the set of that ball without the point zero, then the boundary will be the unision of ##|z|=1## and ##|z|=0##
Thanks though
 
I suspect that "someone which you trust" will get the same unsatisfactory mark at an exam as you :)
 
wrobel said:
I suspect that "someone which you trust" will get the same unsatisfactory mark at an exam as you :)
They're not wrong though, op just phrased it confusingly. Op is asking if you remove the point 0 from the ball, is it now a boundary point? The answer is yes it is.
 
In a metric space, the boundary of a set S is the set of points at distance 0 from the set.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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