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• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 6.1. Limits
• 6.2. Continuous Functions
• 6.3. Discontinuous Functions
• 6.4. Topology and Continuity
• 6.5. Differentiable Functions
• 6.6. A Function Primer
• 7. The Integral
• 8. Sequences of Functions
• 9. Historical Tidbits
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Proposition 6.4.1: Continuity and Topology

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Let f be a function with domain D in R. Then the following statements are equivalent:

1. f is continuous
2. If D is open, then the inverse image of every open set under f is again open.
3. If D is open, then the inverse image of every open interval under f is again open.
4. If D is closed, then the inverse image of every closed set under f is again closed.
5. If D is closed, then the inverse image of every closed interval under f is again closed.
6. The inverse image of every open set under f is the intersection of D with an open set.
7. The inverse image of every closed set under f is the intersection of D with a closed set.

Context

Proof:

(1) => (2): Assume that f is continuous on an open set D. Let U be an open set in the range of f. We need to show that f^-1(U) D is again open. Take any x₀ f^-1(U). That is equivalent to saying that f( x₀) U. Since U is open, we can find an > 0 such that the - neighborhood of f( x₀) is contained in U. For this fixed we can use the continuity of f to pick a > 0 such that

• if | x - x₀| < then | f(x) - f( x₀) | <

This implies that the - neighborhood of x₀is contained in f^-1(U). Hence, the inverse image of the arbitrary set U is open.

(2) => (1): Assume that the inverse image f^-1(U) of every open set U is open. Take any point x₀ U and pick an > 0. Then the - neighborhood of f( x₀) is an open set, so that it's inverse image is again open. That inverse image contains x₀, and since it is open it contains a - neighborhood of x₀for some > 0. But that is exactly what we want:

• if | x - x₀| < then this set is contained in the set f^-1( f( x₀) - , f( x₀) + )

or in other words

• if | x - x₀| < then | f(x) - f( x₀) | <

(2) <=> (3): This follows immediately from the fact that every open set in the real line can be written as the countable union of open intervals.

(2) <=> (4): This follows immediately by looking at complements, i.e. from the fact that

• f^-1( comp(U) ) = comp( f^-1(U) )

That equality should be proved as an exercise.

(4) <=> (5): This follows again by combining the two previous remarks.

(6), (7): This proof is very similar to the proof of (2) <=> (1). In fact, where in that previous proof have we used the fact that the domain D of the function is open ? The details are left as an exercise again.