Fun Exercise (Topology): Show that convergent sequences in Hausdorff spaces have unique limits.

Question

kainui · Answer

I'll try if you give me some hints. I know(?) that Hausdorff means you can create an open ball around two separate points and their intersection is empty or something like this. But that's really about the extent of my knowledge because Hausdorff is supposedly common. I think they also call it T2 which I understand to be something like increasing levels of separation between points. I've recently been self teaching myself some of this and this was stated the other day in the lecture I watched.

across · Answer

Everything you said is correct. :-) The proof of this falls straight out of the definition of Hausdorff: let $\left\{x_n\right\}_{n=1}^\infty$ be a sequence in such a space that converges to two distinct points $a$ and $b$, say. Apply the Hausdorff condition to these two points and the proof should become evident. :-)

across · Answer

attachment

taiofchi · Answer

Let T=(S,τ) be a Hausdorff space.

Let ⟨xn⟩ be a convergent sequence in T.


Then ⟨xn⟩ has exactly one limit.

taiofchi · Answer

From the definition of convergent sequence, we have that ⟨xn⟩ converges to at least one limit.


Suppose limn→∞xn=l and limn→∞xn=m such that l≠m.

As T is Hausdorff, ∃U∈τ:l∈U and ∃V∈τ:m∈V such that U∩V=∅.


Then, from the definition of convergent sequence:

∃NU∈R:  		n>NU	⟹				xn∈U				
∃NV∈R:  		n>NV	⟹				xn∈V				

Taking N=max{NU,NV} we then have:

∃N∈R:n>N⟹xn∈U,xn∈V
But U∩V=∅.


From that contradiction we can see that there can be no such two distinct l and m.

Hence the result.

■