Mixing
How can i prove that? Connected components and continuous function
$begingroup$
$X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)
Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.
Please, help me understand how i can do it. And please excuse my English.
general-topology algebraic-topology
edited Jan 16 at 14:08
Tyrone
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
$endgroup$
add a comment |
$begingroup$
$X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)
Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.
Please, help me understand how i can do it. And please excuse my English.
general-topology algebraic-topology
edited Jan 16 at 14:08
Tyrone
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
$endgroup$
add a comment |
$begingroup$
$X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)
Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.
Please, help me understand how i can do it. And please excuse my English.
general-topology algebraic-topology
edited Jan 16 at 14:08
Tyrone
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
$endgroup$
$X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)
Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.
Please, help me understand how i can do it. And please excuse my English.
general-topology algebraic-topology
general-topology algebraic-topology
edited Jan 16 at 14:08
Tyrone
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
edited Jan 16 at 14:08
Tyrone
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
edited Jan 16 at 14:08
Tyrone
4,80511225
edited Jan 16 at 14:08
Tyrone
4,80511225
edited Jan 16 at 14:08
Tyrone
4,80511225
4,80511225
asked Jan 16 at 13:34
pentarrapentarra
31
asked Jan 16 at 13:34
pentarrapentarra
31
asked Jan 16 at 13:34
pentarrapentarra
31
31
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.
Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.
Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
$endgroup$
add a comment |
$begingroup$
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.
Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
$endgroup$
add a comment |
$begingroup$
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.
Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
$endgroup$
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.
Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
answered Jan 16 at 22:19
Paul FrostPaul Frost
11.1k3934
11.1k3934
add a comment |
add a comment |
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