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A LITTLE ON THE NOTION OF CONTINUITY
TMA4100, MATEMATICS 1, NTNU

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Let a ∈ I
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We say that f is continuous if f is continuous at each point in its domain of definition
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x→a

An alternative way of thinking about the continuity of f , and which also gives a more general
point of view, is as follows: The function f is continuous at a if and only if for each open interval
J containing f (a), the inverse image f −1 (J) contains an open interval I0 containing a
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Alternatively: A is open if A is a union of (disjoint) open intervals
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S UMS , PRODUCTS , FRACTIONS , AND COMPOSITIONS OF CONTINUOUS FUNCTIONS
From the rules of limits, the following proposition follows:
Proposition 2
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Let I, a, and f be as above, and suppose that f is continuous at a
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Then the following holds:
i) f + g is continuous at a
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iii) fg is continuous at a provided g(a) 6= 0
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We prove ii); i) and iii) are proven analogously
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x→a

x→a

x→a

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TMA4100, MATEMATICS 1, NTNU

For illustrative purposes, we here also include a proof using the , δ-definition
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We need to show that there exists a δ > 0 such that
|x − a| < δ =⇒ |f g(x) − f g(a)| < 
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Let us assume that g(a) 6= 0; the converse case is easier (and left as an exercise to the reader)
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Consequently, if |x − a| < δ1 , we have




|f g(x) − f g(a)| ≤
+ |f (a)| |g(x) − g(a)| +
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Hence, we can let δ = min{δ1 , δ2 }
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This completes the proof (of ii))
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It is also a very important result that composing continuous
functions results in continuous functions
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2
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Suppose
moreover that g is an R-valued function defined on some Dg containing f (I) (the image of f ),
and also that g is continuous at f (a)
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Proof
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Proof 1
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Using limits, and the continuity of g and f , we have
lim h(x) = lim g(f (x)) = g(lim f (x)) = g(f (a)) = h(a),

x→a

x→a

x→a

which shows that h is continuous at a
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A LITTLE ON THE NOTION OF CONTINUITY

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Proof 2
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We show that there is δ > 0 such that
|x − a| < δ =⇒ |g(f (x)) − g(f (a))| < 
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In particular, considering y = f (x), we have that
|f (x) − f (a)| < η =⇒ |g(f (x)) − g(f (a))| < 
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But this now follows immediately from the continuity of f
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Let h = g ◦ f , and let J be an open interval containing h(a)
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Let y ∈ h(I)
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Thus, h−1 (J) = f −1 (g −1 (J))
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Hence,
h−1 (J) = f −1 (g −1 (J)) ⊇ f −1 (J0 )
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That is, we
have found an open interval I0 3 a such that
h−1 (J) ⊇ I0 ,


which is what we wanted
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T HE INTERMEDIATE VALUE THEOREM

Intuitively, if f is continuous, its graph should be connected in the sense that if at some point
the graph is below the x-axis, and at some other point, the graph is above the x-axis, then at some
point it must have intersected the x-axis
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1
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Then
there exists c ∈ (a, b) such that f (c) = 0
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Here is the idea: Since [a, b] is connected, being an interval, and f is continuous, it follows that f ([a, b]) must also be connected
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Since f (a) < 0 and f (b) > 0, and this is an interval, it follows
that 0 ∈ [f (a), f (b)] ⊆ f ([a, b])
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Some might be sceptical to this proof, and with good reason; after all, there are several assertion within the above which we have yet to justify
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Let
S := {x ∈ [a, b] : f (x) < 0}
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Let c = sup S
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Suppose first that f (c) < 0
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Since f is continuous, there is δ > 0 such that
|x − c| < δ =⇒ |f (x) − f (c)| < −γf (c)
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Now, consider γ = 21
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This
contradicts that c is an upper bound for S
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By similar reasoning then,
since f is continuous at c, there is δ 0 > 0 such that
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|x − c| < δ 0 =⇒ |f (x) − f (c)| < f (c)
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But this upper bound is smaller than c, which is impossible, since c is the supremum of S
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Thus, f (c) ≥ 0 and f (c) ≤ 0, so the only possibility is that
f (c) = 0
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Then there exists c ∈ (a, b) such that f (c) = g(c)
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Consider h = f − g
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Thus, by the
intermediate value theorem, there is c ∈ (a, b) such that h(c) = 0
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