Question

Let W = {a,b,c,d). Decide whether each set of ordered pairs is a function from Winto W.
(a) (b. a)
(b) {(d, d),
(c,d). (d.a), (c,d)
(c,d) (a, d)) (C) ((a, b),
(c) f(a, b), (b.b), (c,d). (d.b))
(c.a). (a, b). (d, b)]
(d) {(a,a), (b.a),
(b, a), (a, b), (c,d))​

Answer 1

Answer: The answers for each part are before the parts question refer that.

Step-by-step explanation:

Let W = {a,b,c,d). Decide whether each set of ordered pairs is a function from Winto W.

(a) (b. a)

(b) {(d, d),

(c,d). (d.a), (c,d)

(c,d) (a, d)) (C) ((a, b),

(c) f(a, b), (b.b), (c,d). (d.b))

(c.a). (a, b). (d, b)]

(d) {(a,a), (b.a),

(b, a), (a, b), (c,d))​

Conditions for a function have to be satisfied that are:-

For the first question, I have come to a seemingly necessary and sufficient condition on K, which may be a bit too complicated : K is the support of a continuous function iff for all x∈K, for all neighborhood V of x, there exists a nonempty open set U⊂K∩V. The fact that it is necessary is pretty obvious, and for the sufficient aspect, I have considered the connected components of such a K, which are segments. I call a ''trivial segment'' a segment which is a singleton. Putting aside the trivial case where K is empty, there is at least one nontrivial segment among the connected components of K. The following two cases can arise:

If the number of nontrivial connected components is finite, there is no trivial segment and f can be defined as a triangle-shape function on all the connected components and 0 otherwise. f is continuous and has support K.

If the number of nontrivial connected components is infinite, it is countable and the nontrivial components can be ordered as {Cn}n>0. On each Cn f is defined as a triangle-shape function of height 1/n, and 0 outside of ∪n>0Cn. Again, f is continuous and has support K.

I am now considering the extension to other spaces and I can't really figure out a way to generalize the result. The necessary condition on K seems to remain valid on a general topological space, but for the other part of the proof, I have used the particular nature of the connected sets of R, which cannot be used in the general case. Is there a way to claim a more general result, maybe involving a simpler characterization?:-

So the set has to be one-one and onto that makes it a function checking that for all the parts clearly gives the:-

Answer:-a)yes

b)no

c)Yes

d)no

e)yes

f)no

g)no

h)yes.

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