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# Prove Set Is Open Given Continuous Function

An equivalent condition for a continuous function is that the inverse image of any open set is open. Now \$(0,infty)subsetmathbb R\$ is open. Hence \$xmid f(x)gt0=f^-1(0,infty)\$ is open.

â€¢ Other Related Knowledge ofa continuous function

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Conditions on \$X\$ ensuring that a non-constant continuous function \$f: Xto mathbbC\$ exists

Just to give this an answer:As has been pointed out in the comments, \$|X|>1\$ is both necessary and sufficient: if \$x,yin X\$ are distinct, Urysohn's lemma ensures that there is a continuous function \$f:Xto,\$ such that \$f(x)=0\$ and \$f(y)=1\$

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\$R/I\$ when \$R\$ is the ring of real continuous functions

Hint: What is the kernel of the ring homomorphism \$Rto mathbb R\$, \$fmapsto f(0.5)\$?

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Determining whether a given set can be an image of continuous function

The second and third contradict the intermediate value theorem.For the first one, i will give you an example with image \$(0,1]\$: \$x mapsto frac1x^21\$. I think you are capable of adjusting this to \$[0,1). \$.

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\$epsilon-delta\$ definition of continuous functions

How about a particular proof instead? Let's show that \$f(x) = x^2\$ is uniformly continuous on \$,\$. Fix \$epsilon > 0\$. Observe that \$\$ |f(x) - f(y)| = |x^2 - y^2| = |x - y||x y| \$\$ Now, if \$x,y in ,\$, then \$|xy| leq 2\$. Thus, \$\$ |f(x) - f(y)| leq 2|x - y| \$\$ Now, choose \$delta = fracepsilon2 > 0\$. If \$x,y in ,\$ and \$|x - y|

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Function always continuous in a Sobolev Space?

Your reasoning is correct. If \$f\$ cannot be redefined on a set of measure zero to become a continuous function, then \$f\$ is not in \$H^1(-1,1)\$. Chances are that sooner or later you will encounter Sobolev functions of more than one variable. Those need not have a continuous representative. For example, \$\$ f(x,y) := sqrt- log(x^2y^2) \$\$ is in \$H^1(D)\$ where \$D\$ is the disk in two dimensions. It cannot be made continuous, or even bounded, by changing its values on a set of measure zero.

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\$X\$ is simply connected iff every continuous \$f:S^1

ightarrow X\$ has a continuous extension to the unit disc?

So indeed \$F(s,1)=barF(Phi(s,1))=barF(e^2pi is)=barsigma(e^2pi is)=sigma(s)\$.But now you need to show that \$F(s,0)=F(0,t)=F(1,t)=x_0\$. So we calculate: \$F(s,0)=barF(0)\$ (meaning \$F\$ is a free homotopy between \$sigma\$ and constant \$barF(0)\$) and \$F(0,t)=F(1,t)=barF(t)\$ (which is not quite enough for it to be a path homotopy). Note that \$tin,\$ and we treat the interval \$,\$ as a subset of the complex plane \$mathbbC\$, formally \$,times0\$.Warning: be aware that in general the existance of a free homotopy need not imply the existance of a based homotopy. We need to do some work in this particular case. So the question is: can we choose \$barF\$ in such a way that \$barF(t)=x_0\$ for all \$tin,subseteqmathbbC\$? We know that \$barF(1)=x_0\$ because \$barsigma(1)=sigma(0)=x_0\$.So lets ask another question: does there exist a continuous function \$Psi:Dto D\$ such that \$Psi(z)=z\$ for \$zin S^1\$ and \$Psi(t)=1\$ for \$tin,\$? If such function exists then you simply modify \$barF\$ by taking \$barFcircPsi\$.And indeed since \$S^1cup,\$ is closed and \$Psi\$ restricted to that subset is continuous then such function exists by the (generalized) Tietze extension theorem (together with the fact that \$D\$ is a retract of \$mathbbR^2\$). There might be an explicit formula but I just could not figure it out

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When is the inverse image of continuous function a submanifold

You can not really say much - given an arbitrary closed set \$A subseteq mathbbR^n\$, the distance function \$f(x) = d(A,x)\$ is continuous and \$A = f^-1(0)\$ so any closed subset is the inverse image of a continuous function but most of them are very far from being manifolds

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Is an equicontinuous family of uniformly continuous functions necessarily uniformly equicontinuous?

Let \$F\$ be the family of differentiable functions \$f:mathbb Rtomathbb R\$ such that(i) \$f'\$ is bounded(ii) \$|f'(t)|le |t|\$ for all \$t\$.(The bound in (i) is allowed to depend on \$f\$.)Then (i) shows every \$fin F\$ is uniformly continuous, and (ii) shows that \$F\$ is (pointwise) equicontinuous, but \$F\$ is not uniformly equicontinuous. (For example, consider \$f_n\$ with \$f_n'(t)=min(n, |t|)\$: If \$s,t>n\$ then \$|f_n(s)-f_n(t)|=n|s-t|\$.)

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