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In which of the following cases there exist a one one continuous function from S to T, [closed]
In which of the following cases there exist a one one continuous function from $S$ to $T$,
(a) $S=[0,1]$ to $T=(0,1)$
(b) $S=Bbb R$ to $T=[0,1]$
(c) $S=Bbb R$ to $T=(0,1)$
(d) $S=(0,1]$ to $T=(0,1)$
(a)I think a) false since continuous function preserves compactness but what about others dont know . Please someone explain.
real-analysis
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
closed as off-topic by Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos Nov 22 '18 at 8:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
In which of the following cases there exist a one one continuous function from $S$ to $T$,
(a) $S=[0,1]$ to $T=(0,1)$
(b) $S=Bbb R$ to $T=[0,1]$
(c) $S=Bbb R$ to $T=(0,1)$
(d) $S=(0,1]$ to $T=(0,1)$
(a)I think a) false since continuous function preserves compactness but what about others dont know . Please someone explain.
real-analysis
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
closed as off-topic by Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos Nov 22 '18 at 8:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
In which of the following cases there exist a one one continuous function from $S$ to $T$,
(a) $S=[0,1]$ to $T=(0,1)$
(b) $S=Bbb R$ to $T=[0,1]$
(c) $S=Bbb R$ to $T=(0,1)$
(d) $S=(0,1]$ to $T=(0,1)$
(a)I think a) false since continuous function preserves compactness but what about others dont know . Please someone explain.
real-analysis
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
In which of the following cases there exist a one one continuous function from $S$ to $T$,
(a) $S=[0,1]$ to $T=(0,1)$
(b) $S=Bbb R$ to $T=[0,1]$
(c) $S=Bbb R$ to $T=(0,1)$
(d) $S=(0,1]$ to $T=(0,1)$
(a)I think a) false since continuous function preserves compactness but what about others dont know . Please someone explain.
real-analysis
real-analysis
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
edited Nov 22 '18 at 2:07
John Nash
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
asked Nov 22 '18 at 1:51
John NashJohn Nash
7418
7418
closed as off-topic by Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos Nov 22 '18 at 8:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos Nov 22 '18 at 8:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Steve Kass, user10354138, A. Pongrácz, Kavi Rama Murthy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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Such a function exists in all four cases. The key idea is that any interval in the real numbers can be continuously mapped into any other interval just by stretching or squashing and then shifting it over.
For example, in (a), define $f$ by $f(x) = x / 2 + 1/4$. Check that this is continuous as a function $Bbb{R} to Bbb{R}$, that $f(x) in T$ whenever $x in S$, and that $f$ is one-to-one.
(Note that $f$ is not onto. It is true that because $S$ is compact, the image $f(S)$ must also be compact, but the image is not all of $T$, so the argument you sketched does not work. It's fine for $f$ to map $S$ to a compact subspace of $T$ even though $T$ is not compact.)
In (b), let's start by defining a function $g: Bbb{R} to Bbb{R}$ whose image is bounded. Define $g(x) = e^x$ (for $x le 0$), $g(x) = 2 - e^{-x}$ (for $x ge 0$). This is clearly continuous on the left side $x le 0$ and on the right side $x ge 0$, and because the two sides agree at $0$, it's continuous on $Bbb{R}$. The image of $g$ is $(0, 2)$, and $g$ is strictly increasing and therefore one-to-one. Now you can easily modify $g$ so that its image is inside $[0, 1]$.
(c) and (d) are similar, and I'll leave them for you.
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Such a function exists in all four cases. The key idea is that any interval in the real numbers can be continuously mapped into any other interval just by stretching or squashing and then shifting it over.
For example, in (a), define $f$ by $f(x) = x / 2 + 1/4$. Check that this is continuous as a function $Bbb{R} to Bbb{R}$, that $f(x) in T$ whenever $x in S$, and that $f$ is one-to-one.
(Note that $f$ is not onto. It is true that because $S$ is compact, the image $f(S)$ must also be compact, but the image is not all of $T$, so the argument you sketched does not work. It's fine for $f$ to map $S$ to a compact subspace of $T$ even though $T$ is not compact.)
In (b), let's start by defining a function $g: Bbb{R} to Bbb{R}$ whose image is bounded. Define $g(x) = e^x$ (for $x le 0$), $g(x) = 2 - e^{-x}$ (for $x ge 0$). This is clearly continuous on the left side $x le 0$ and on the right side $x ge 0$, and because the two sides agree at $0$, it's continuous on $Bbb{R}$. The image of $g$ is $(0, 2)$, and $g$ is strictly increasing and therefore one-to-one. Now you can easily modify $g$ so that its image is inside $[0, 1]$.
(c) and (d) are similar, and I'll leave them for you.
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
add a comment |
Such a function exists in all four cases. The key idea is that any interval in the real numbers can be continuously mapped into any other interval just by stretching or squashing and then shifting it over.
For example, in (a), define $f$ by $f(x) = x / 2 + 1/4$. Check that this is continuous as a function $Bbb{R} to Bbb{R}$, that $f(x) in T$ whenever $x in S$, and that $f$ is one-to-one.
(Note that $f$ is not onto. It is true that because $S$ is compact, the image $f(S)$ must also be compact, but the image is not all of $T$, so the argument you sketched does not work. It's fine for $f$ to map $S$ to a compact subspace of $T$ even though $T$ is not compact.)
In (b), let's start by defining a function $g: Bbb{R} to Bbb{R}$ whose image is bounded. Define $g(x) = e^x$ (for $x le 0$), $g(x) = 2 - e^{-x}$ (for $x ge 0$). This is clearly continuous on the left side $x le 0$ and on the right side $x ge 0$, and because the two sides agree at $0$, it's continuous on $Bbb{R}$. The image of $g$ is $(0, 2)$, and $g$ is strictly increasing and therefore one-to-one. Now you can easily modify $g$ so that its image is inside $[0, 1]$.
(c) and (d) are similar, and I'll leave them for you.
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
add a comment |
Such a function exists in all four cases. The key idea is that any interval in the real numbers can be continuously mapped into any other interval just by stretching or squashing and then shifting it over.
For example, in (a), define $f$ by $f(x) = x / 2 + 1/4$. Check that this is continuous as a function $Bbb{R} to Bbb{R}$, that $f(x) in T$ whenever $x in S$, and that $f$ is one-to-one.
(Note that $f$ is not onto. It is true that because $S$ is compact, the image $f(S)$ must also be compact, but the image is not all of $T$, so the argument you sketched does not work. It's fine for $f$ to map $S$ to a compact subspace of $T$ even though $T$ is not compact.)
In (b), let's start by defining a function $g: Bbb{R} to Bbb{R}$ whose image is bounded. Define $g(x) = e^x$ (for $x le 0$), $g(x) = 2 - e^{-x}$ (for $x ge 0$). This is clearly continuous on the left side $x le 0$ and on the right side $x ge 0$, and because the two sides agree at $0$, it's continuous on $Bbb{R}$. The image of $g$ is $(0, 2)$, and $g$ is strictly increasing and therefore one-to-one. Now you can easily modify $g$ so that its image is inside $[0, 1]$.
(c) and (d) are similar, and I'll leave them for you.
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
Such a function exists in all four cases. The key idea is that any interval in the real numbers can be continuously mapped into any other interval just by stretching or squashing and then shifting it over.
For example, in (a), define $f$ by $f(x) = x / 2 + 1/4$. Check that this is continuous as a function $Bbb{R} to Bbb{R}$, that $f(x) in T$ whenever $x in S$, and that $f$ is one-to-one.
(Note that $f$ is not onto. It is true that because $S$ is compact, the image $f(S)$ must also be compact, but the image is not all of $T$, so the argument you sketched does not work. It's fine for $f$ to map $S$ to a compact subspace of $T$ even though $T$ is not compact.)
In (b), let's start by defining a function $g: Bbb{R} to Bbb{R}$ whose image is bounded. Define $g(x) = e^x$ (for $x le 0$), $g(x) = 2 - e^{-x}$ (for $x ge 0$). This is clearly continuous on the left side $x le 0$ and on the right side $x ge 0$, and because the two sides agree at $0$, it's continuous on $Bbb{R}$. The image of $g$ is $(0, 2)$, and $g$ is strictly increasing and therefore one-to-one. Now you can easily modify $g$ so that its image is inside $[0, 1]$.
(c) and (d) are similar, and I'll leave them for you.
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
answered Nov 22 '18 at 5:14
Hew WolffHew Wolff
2,240716
2,240716
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
add a comment |
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
You mean in order to check one one compactness not require
– John Nash
Nov 22 '18 at 9:06
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
Whether the spaces are compact and whether the function is one-to-one are two different questions, and they have nothing to do with each other. If the function is required to be onto as well as one-to-one (that is, the function is a bijection), then you might be able to say something about the compactness of the spaces.
– Hew Wolff
Nov 23 '18 at 21:28
add a comment |
