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  1. Consider the map $f\colon(\mathbb{R}^2,\tau)\to (\mathbb{R},J)$ given by $f(x,y)=x+y$, where $\tau$ is the standard topology on $\mathbb{R}^2$ and $J$ is the order topology on $\mathbb{R}$. Find and sketch on the same plane:

    • (a) The image of $\{0\}$ under inverse of $f$.
    • (b) The image of $[0,1)$ under inverse of $f$.
    • (c) Is $f$ continuous? Why?
  2. In $\mathbb{R}^2$, define a map $d\colon\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ by $d(x,y)=|x_1-y_1|+|x_2-y_2|$ where $x=(x_1,x_2)$, $y=(y_1,y_2)$.

    • (a) Show that $d$ is a metric on $\mathbb{R}^2$.
    • (b) Find the basis for the topology on $\mathbb{R}^2$ induced by $d$.
    • (c) How is the topology induced by $d$ related to the standard topology on $\mathbb{R}^2$?
  3. Let $S$ denote the unit circle $S=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2=1\}$, considered as a subspace of the plane $\mathbb{R}^2$, and let $f\colon [0,1)\to S$ be the map defined $by f(t)=(\cos(2\pi t),\sin(2\pi t))$. Show that $f$ is continuous but not a homeomorphism.

  4. Show that the function $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x+1$ is a homeomorphism.

My attempt.

Question 1. (a) $f^{-1}(\{0\}) = \{ (x,y)\in\mathbb{R}^2\mid f(x,y)=0\} = \{\text{all points in the line } x=-y\}$.

Question 2. (a) $d(x,y)=|x_1-y_1|+|x_2-y_2|=|y_1-x_1|+|y_2-x_2|=d(y,x)$ $$\begin{align*} d(x,y)&=|x_1-y_1|+|x_2-y_2|=|x_1-z_1 + z_1-y_1|+|x_2-z_2 +z_2-y_2|\\ &\leq |x_1-z_1|+|x_2-z_2|+|z_1-y_1|+|z_2-y_2|\\ &\leq d(x,z)+d(z,y) \end{align*}$$ therefore $d$ is a metric on $\mathbb{R}^2$.

Question 4. $f(1)=4$, $f(2)=7$ so $f$ is one to one hence bijection, I think $f$ carries a basis element from $\mathbb{R}$ to $\mathbb{R}$ and vice versa, hence $f$ is automatically homeomorphism

I wish to know the correctness of my attempt , and how I can attempt Question 1(b),(c); Question 2(b),(c); and Question 3.

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You can use latex (between dollar signs) on this site to format your question. It will become more readable for everybody, and you will get better answers. – Thomas Rot May 6 '11 at 15:39
Did you prove that on $\mathbb R$ the order topology coincides with the standard topology? – Asaf Karagila May 6 '11 at 16:03
no i didnt check it! – temba alloyce May 6 '11 at 16:19
up vote 1 down vote accepted

You can answer 1b very similarly to how you answered 1a. For 1c, refer to the definition of continuity.

For 2a, don't forget to show that the metric is positive-definite! For 2b and 2c, you need to think about which sets in $\mathbb{R}^2$ are open in the new topology.

For Q4, you should prove your claim that "$f$ carries a basis element from $\mathbb{R}$ to $\mathbb{R}$." By "vice-versa," do you mean that $f^{-1}$ also carries basis elements to basis elements? You need to prove that, too. I think your argument works, but you should double-check your definition of homeomorphism again. The key is that $f^{-1}$ carries base sets to base sets; if you're going to examine $f$ and $f^{-1}$, it might not matter much, but you should double-check anyway.

Q3, as you've written it, doesn't make sense; do you mean "show that $f$ is continuous but not a homeomorph*ism*"? Just show that $f$ satisfies the definition from Q4.

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its to show that f is not homeomorphism!, i have no idea – temba alloyce May 6 '11 at 16:24
Sorry, I should have said, "show that $f$ DOESN'T satisfy the definition from Q4"! – Adam Saltz May 6 '11 at 22:12

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