S18 - Munkres 18 Ex 18.1(Morten Poulsen Recall the-denition...

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September 29, 2008 Munkres § 18 Ex. 18.1 (Morten Poulsen). Recall the ε - δ -deﬁnition of continuity: A function f : R R is said to be continuous if a R ε R + δ R + x R : | x - a | < δ ⇒ | f ( x ) - f ( a ) | < ε. Let T be the standard topology on R generated by the open intervals. Theorem 1. For functions f : R R the following are equivalent: ( i ) a R ε R + δ R + x R : | x - a | < δ ⇒ | f ( x ) - f ( a ) | < ε . ( ii ) a R ε R + δ R + : f (( a - δ,a + δ )) ( f ( a ) - ε,f ( a ) + ε ) . ( iii ) U ∈ T : f - 1 ( U ) ∈ T . Proof. “( i ) ( ii )”: Clear. “( ii ) ( iii )”: Let U ∈ T . If a f - 1 ( U ) then f ( a ) U . Since U is open there exists ε R + such that ( f ( a ) - ε,f ( a ) + ε ) U . By assumption there exists δ a R + such that f (( a - δ a ,a + δ a )) ( f ( a ) - ε,f ( a ) + ε ), hence ( a - δ a ,a + δ a ) f - 1 ( U ). It follows that f - 1 ( U ) = S a f - 1 ( U ) ( a - δ a ,a + δ a ), i.e. open. “( iii ) ( ii )”: Let a R . Given ε R + then f - 1 (( f ( a ) - ε,f ( a ) + ε )) is open and contains a . Hence there exists δ R + such that ( a - δ,a + δ ) f - 1 (( f ( a ) - ε,f ( a ) + ε )). It follows that f (( a - δ,a + δ )) ( f ( a ) - ε,f ( a ) + ε ). ± Ex. 18.2. Let f : R → { 0 } be the constant map. Then 2004 is a limit point of R but f (2004) = 0

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S18 - Munkres 18 Ex 18.1(Morten Poulsen Recall the-denition...

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