Let S=(0,1) cup(1,2) cup(3,4) and T={0,1,2,3} | vedclass

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(A) Given $S = (0,1) \cup (1,2) \cup (3,4)$ and $T = \{0, 1, 2, 3\}$.$(A)$ Since $S$ is an infinite set and $T$ has $4$ elements,there are $4^{|S|}$ f

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$A, C, D$

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(A) Given $S = (0,1) \cup (1,2) \cup (3,4)$ and $T = \{0, 1, 2, 3\}$.$(A)$ Since $S$ is an infinite set and $T$ has $4$ elements,there are $4^{|S|}$ functions from $S$ to $T$. Since $|S|$ is infinite,there are infinitely many functions. Thus,$(A)$ is true.$(B)$ $A$ strictly increasing function $f: S 	o T$ must map each connected component of $S$ to a single value because the image of a connected set under a continuous function is connected. However,$T$ is a discrete set. For $f$ to be strictly increasing,it must be constant on each component. Since $T$ is finite,there are only finitely many such functions. Thus,$(B)$ is false.$(C)$ $A$ continuous function $f: S 	o T$ must be constant on each connected component of $S$ because the image of a connected set $(0,1)$,$(1,2)$,and $(3,4)$ under $f$ must be a connected subset of $T$. The only connected subsets of $T$ are singleton sets. Thus,$f$ must take a constant value on each of the $3$ components. There are $4$ choices for each component,so there are $4 	imes 4 	imes 4 = 64$ such functions. Since $64 \le 120$,$(C)$ is true.$(D)$ Any continuous function $f: S 	o T$ is a constant function on each connected component of $S$. $A$ constant function is differentiable everywhere on its domain. Thus,$(D)$ is true.Therefore,the correct statements are $(A)$,$(C)$,and $(D)$.

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