Difference between revisions of "Continuity definitions are equivalent"
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| + | ==Statement== | ||
| + | The definitions of continuity for a function <math>f:(X,d)\rightarrow(Y,d')</math> from one [[Metric space|metric space]] to another is the same as <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> being continuous (where the topologies are those [[Topology induced by a metric|induced by the metric]] are the same, that is | ||
| + | # <math>\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))</math> | ||
| + | # <math>\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}</math> | ||
| − | {{Theorem|Topology}} | + | ==Proof== |
| + | ===<math>\implies</math>=== | ||
| + | Suppose <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> is continuous. | ||
| + | |||
| + | Let <math>V\in\mathcal{K}</math> - that is <math>V</math> is [[Open set|open]] within <math>Y</math> | ||
| + | |||
| + | Let <math>x\in f^{-1}(V)</math> be given. | ||
| + | |||
| + | Then because <math>V</math> is open, <math>\exists\epsilon>0</math> such that <math>B_\epsilon(f(x))\subset V</math> (note that <math>f(x)\in V</math> by definition of where we choose x from). | ||
| + | |||
| + | But by continuity of <math>f</math> we know that <math>\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V</math> | ||
| + | |||
| + | Thus <math>B_\delta(x)\subset f^{-1}(V)</math> (as for all <math>a</math> in the ball, the thing <math>f</math> maps it to is in the ball of radius <math>\epsilon</math> about <math>f(x)</math>). | ||
| + | |||
| + | Since <math>x</math> was arbitrary we have <math>\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)</math>, thus <math>f^{-1}(V)</math> is open. | ||
| + | |||
| + | ===<math>\impliedby</math>=== | ||
| + | Choose any <math>x\in X</math> | ||
| + | |||
| + | Let <math>\epsilon>0</math> be given. | ||
| + | |||
| + | As <math>B_\epsilon(f(x))</math> is an open set, the hypothesis implies that <math>f^{-1}(B_\epsilon(f(x)))</math> is open in <math>X</math> | ||
| + | |||
| + | Since <math>x\in f^{-1}(B_\epsilon(f(x)))</math> and <math>f^{-1}(B_\epsilon(f(x)))</math> is open, it is a neighborhood to all of its points, that means | ||
| + | |||
| + | <math>\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math> | ||
| + | |||
| + | ''Note:'' we have now shown that <math>\forall\epsilon>0\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math> | ||
| + | |||
| + | Using the [[Implies and subset relation|implies and subset relation]] we see <math>a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))</math> | ||
| + | |||
| + | Or just <math>a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))</math> | ||
| + | |||
| + | Thus it is continuous at <math>x</math>, since <math>x</math> was arbitrary, it is continuous. | ||
| + | |||
| + | {{Theorem Of|Topology}} | ||
Latest revision as of 07:20, 27 April 2015
Contents
[hide]Statement
The definitions of continuity for a function f:(X,d)→(Y,d′) from one metric space to another is the same as f:(X,J)→(Y,K) being continuous (where the topologies are those induced by the metric are the same, that is
- ∀a∈X∀ϵ>0∃δ>0:x∈Bδ(a)⟹f(x)∈Bϵ(f(a))
- ∀V∈K:f−1(V)∈J
Proof
⟹
Suppose f:(X,J)→(Y,K) is continuous.
Let V∈K - that is V is open within Y
Let x∈f−1(V) be given.
Then because V is open, ∃ϵ>0 such that Bϵ(f(x))⊂V (note that f(x)∈V by definition of where we choose x from).
But by continuity of f we know that ∃δ>0:a∈Bδ(x)⟹f(a)∈Bϵ(f(x))⊂V
Thus Bδ(x)⊂f−1(V) (as for all a in the ball, the thing f maps it to is in the ball of radius ϵ about f(x)).
Since x was arbitrary we have ∀x∈f−1(V)∃an open ball containing x⊂f−1(V), thus f−1(V) is open.
⟸
Choose any x∈X
Let ϵ>0 be given.
As Bϵ(f(x)) is an open set, the hypothesis implies that f−1(Bϵ(f(x))) is open in X
Since x∈f−1(Bϵ(f(x))) and f−1(Bϵ(f(x))) is open, it is a neighborhood to all of its points, that means
∃δ>0:Bδ(x)⊂f−1(Bϵ(f(x)))
Note: we have now shown that ∀ϵ>0∃δ>0:Bδ(x)⊂f−1(Bϵ(f(x)))
Using the implies and subset relation we see a∈Bδ(x)⟹a∈f−1(Bϵ(f(x))) which then ⟹f(a)∈Bϵ(f(x))
Or just a∈Bδ(x)⟹f(a)∈Bϵ(f(x)))
Thus it is continuous at x, since x was arbitrary, it is continuous.
