Analysis-1-Continuity and Connectedness, guaranteed and verified 100% Pass

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Continuity and Connectedness



Recall that:

Def. Two subsets 𝐴, 𝐵 of a metric space 𝑋, 𝑑 are said to be separated if 𝐴 ∩ 𝐵̅=∅
and 𝐴̅ ∩ 𝐵 = ∅ (i.e., no point of 𝐴 lies in the closure of 𝐵 and no point of 𝐵 lies in
the closure of 𝐴).

Def. A set 𝐸 ⊆ 𝑋, 𝑑 a metric space is said to be connected if 𝐸 is not the union of
two nonempty separated sets.



Ex. if 𝐴 = (0,1) and 𝐵 = (1,2) , then 𝐴 and 𝐵 are separated sets since
𝐴̅ = [0,1], 𝐵̅ = [1,2]
thus: 𝐴 ∩ 𝐵̅ = (0,1) ∩ [1,2] = ∅ and
𝐴̅ ∩ 𝐵 = [0,1] ∩ (1,2) = ∅.
Thus the set 𝐴 ∪ 𝐵 = (0,1) ∪ (1,2) is not a connected set.


Ex. If 𝐴 = (0,1] and 𝐵 = (1,2), then 𝐴 and 𝐵 are not separated since
𝐵̅ = [1,2] and thus

𝐴 ∩ 𝐵̅ = (0,1]∩ [1,2] = {1} ≠ ∅

(notice that 𝐴̅ ∩ 𝐵 = [0,1] ∩ (1,2) = ∅).



Theorem: A subset 𝐸 ⊆ ℝ is connected if and only if , if 𝑥𝜖𝐸, 𝑦𝜖𝐸 and
𝑥 < 𝑧 < 𝑦 then 𝑧𝜖𝐸 .

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Theorem: If 𝑓 is a continuous mapping of a metric space 𝑋 into a metric space 𝑌,
and if 𝐸 is a connected subset of 𝑋 then 𝑓(𝐸 ) is connected.


Proof: (This will be a proof by contradiction) Assume the contrary, i.e. that 𝑓 is a
continuous mapping and 𝑓(𝐸 ) is not connected.

Thus 𝑓(𝐸 ) = 𝐴 ∪ 𝐵, where 𝐴 and 𝐵 are non-empty separated sets.

𝑋 𝑌

𝐸
𝐻 𝐵
𝑓 𝐴
𝐺



Let 𝐺 = 𝐸 ∩ 𝑓 −1 (𝐴), 𝐻 = 𝐸 ∩ 𝑓 −1 (𝐵).

Then 𝐸 = 𝐺 ∪ 𝐻 and neither 𝐺 nor 𝐻 is empty.

Since 𝐴 ⊆ 𝐴̅, we have 𝐺 ⊆ 𝑓 −1 (𝐴̅) and 𝑓 −1 (𝐴̅) is closed because 𝑓 is
continuous and 𝐴̅ is closed (inverse image of a closed set is closed when 𝑓 is
continuous).

Since 𝑓 −1 (𝐴̅) is closed, 𝐺̅ ⊆ 𝑓 −1 (𝐴̅).

This means that 𝑓(𝐺̅ ) ⊆ 𝐴̅.

Since 𝑓 (𝐻 ) = 𝐵 and 𝐴̅ ∩ 𝐵 = ∅ (𝐴 and 𝐵 are separated sets), 𝐺̅ ∩ 𝐻 = ∅.

(If 𝑦𝜖𝐺̅ ∩ 𝐻, then 𝑓(𝑦)𝜖𝐴̅ since 𝑦𝜖𝐺̅ , and 𝑓(𝑦)𝜖𝐵 since 𝑦𝜖𝐻, but 𝐴̅ ∩ 𝐵 = ∅ ).

̅ = ∅.
A similar argument shows 𝐺 ∩ 𝐻

But that would mean that 𝐺, 𝐻 are separated sets with 𝐸 = 𝐺 ∪ 𝐻 and thus 𝐸 is
not connected, a contradiction.

Thus 𝑓(𝐸) is connected.