Use the Lebesgue number lemma to prove that $f$ is uniformly continuous if $M$ is compact and $f$ continuous
Suppose that $M$ is covering compact and that $f: M to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous.
Given $epsilon >0$, the set $mathcal{U} = {B_{epsilon/2}(p) mid p in N}$ is an open covering of $f(M)$. Since $f$ is continuous, $f^{-1}(B_{epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering
$$f(M) subset B_{epsilon/2}(p_{1})cup cdots cup B_{epsilon/2}(p_{n}).$$
Therefore,
$$f^{-1}(B_{epsilon/2}(p_{1}))cup cdots cup f^{-1}(B_{epsilon/2}(p_{n}))$$
is a finite subcovering of $M$. By the Lebesgue nummber lema, there is a $delta>0$ such that for every $x in B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$ for some $i$. So, if $d(x,y) < delta$, the "$y$" are in some $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$
Is this correct?
proof-verification continuity metric-spaces compactness
edited Dec 28 '18 at 20:06
Hayk
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
add a comment |
Suppose that $M$ is covering compact and that $f: M to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous.
Given $epsilon >0$, the set $mathcal{U} = {B_{epsilon/2}(p) mid p in N}$ is an open covering of $f(M)$. Since $f$ is continuous, $f^{-1}(B_{epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering
$$f(M) subset B_{epsilon/2}(p_{1})cup cdots cup B_{epsilon/2}(p_{n}).$$
Therefore,
$$f^{-1}(B_{epsilon/2}(p_{1}))cup cdots cup f^{-1}(B_{epsilon/2}(p_{n}))$$
is a finite subcovering of $M$. By the Lebesgue nummber lema, there is a $delta>0$ such that for every $x in B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$ for some $i$. So, if $d(x,y) < delta$, the "$y$" are in some $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$
Is this correct?
proof-verification continuity metric-spaces compactness
edited Dec 28 '18 at 20:06
Hayk
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
add a comment |
Suppose that $M$ is covering compact and that $f: M to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous.
Given $epsilon >0$, the set $mathcal{U} = {B_{epsilon/2}(p) mid p in N}$ is an open covering of $f(M)$. Since $f$ is continuous, $f^{-1}(B_{epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering
$$f(M) subset B_{epsilon/2}(p_{1})cup cdots cup B_{epsilon/2}(p_{n}).$$
Therefore,
$$f^{-1}(B_{epsilon/2}(p_{1}))cup cdots cup f^{-1}(B_{epsilon/2}(p_{n}))$$
is a finite subcovering of $M$. By the Lebesgue nummber lema, there is a $delta>0$ such that for every $x in B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$ for some $i$. So, if $d(x,y) < delta$, the "$y$" are in some $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$
Is this correct?
proof-verification continuity metric-spaces compactness
edited Dec 28 '18 at 20:06
Hayk
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
Suppose that $M$ is covering compact and that $f: M to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous.
Given $epsilon >0$, the set $mathcal{U} = {B_{epsilon/2}(p) mid p in N}$ is an open covering of $f(M)$. Since $f$ is continuous, $f^{-1}(B_{epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering
$$f(M) subset B_{epsilon/2}(p_{1})cup cdots cup B_{epsilon/2}(p_{n}).$$
Therefore,
$$f^{-1}(B_{epsilon/2}(p_{1}))cup cdots cup f^{-1}(B_{epsilon/2}(p_{n}))$$
is a finite subcovering of $M$. By the Lebesgue nummber lema, there is a $delta>0$ such that for every $x in B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$ for some $i$. So, if $d(x,y) < delta$, the "$y$" are in some $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$
Is this correct?
proof-verification continuity metric-spaces compactness
proof-verification continuity metric-spaces compactness
edited Dec 28 '18 at 20:06
Hayk
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
edited Dec 28 '18 at 20:06
Hayk
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
edited Dec 28 '18 at 20:06
Hayk
2,1171213
edited Dec 28 '18 at 20:06
Hayk
2,1171213
edited Dec 28 '18 at 20:06
Hayk
2,1171213
2,1171213
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
asked Dec 28 '18 at 19:33
Lucas CorrêaLucas Corrêa
1,5521321
1,5521321
add a comment |
add a comment |
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The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.
If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
add a comment |
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1 Answer
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1 Answer
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The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.
If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
add a comment |
The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.
If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
add a comment |
The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.
If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.
If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
answered Dec 28 '18 at 22:55
Henno BrandsmaHenno Brandsma
105k347114
105k347114
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
add a comment |
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
Thank you! Your answer is much clearer
– Lucas Corrêa
Dec 29 '18 at 2:05
add a comment |
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