Confusion arising from the 'infiniteness' of a sequence.
$begingroup$
Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?
My thought process was follows:
If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.
*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.
sequences-and-series infinity
asked Dec 30 '18 at 8:44
KaindKaind
749414
$endgroup$
add a comment |
$begingroup$
Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?
My thought process was follows:
If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.
*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.
sequences-and-series infinity
asked Dec 30 '18 at 8:44
KaindKaind
749414
$endgroup$
add a comment |
$begingroup$
Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?
My thought process was follows:
If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.
*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.
sequences-and-series infinity
asked Dec 30 '18 at 8:44
KaindKaind
749414
$endgroup$
Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?
My thought process was follows:
If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.
*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.
sequences-and-series infinity
sequences-and-series infinity
asked Dec 30 '18 at 8:44
KaindKaind
749414
asked Dec 30 '18 at 8:44
KaindKaind
749414
asked Dec 30 '18 at 8:44
KaindKaind
749414
asked Dec 30 '18 at 8:44
KaindKaind
749414
asked Dec 30 '18 at 8:44
KaindKaind
749414
749414
add a comment |
add a comment |
1 Answer
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Assume that such a sequence $(t_i)_i$ exists.
$T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.
Some $iinmathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
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1 Answer
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1 Answer
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$begingroup$
Assume that such a sequence $(t_i)_i$ exists.
$T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.
Some $iinmathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Assume that such a sequence $(t_i)_i$ exists.
$T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.
Some $iinmathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Assume that such a sequence $(t_i)_i$ exists.
$T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.
Some $iinmathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
$endgroup$
Assume that such a sequence $(t_i)_i$ exists.
$T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.
Some $iinmathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
edited Dec 30 '18 at 9:07
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
answered Dec 30 '18 at 8:58
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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