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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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  • $\begingroup$ 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. $\endgroup$
    – Thomas Rot
    May 6, 2011 at 15:39
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    $\begingroup$ Did you prove that on $\mathbb R$ the order topology coincides with the standard topology? $\endgroup$
    – Asaf Karagila
    May 6, 2011 at 16:03
  • $\begingroup$ no i didnt check it! $\endgroup$ May 6, 2011 at 16:19

1 Answer 1

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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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  • $\begingroup$ its to show that f is not homeomorphism!, i have no idea $\endgroup$ May 6, 2011 at 16:24
  • $\begingroup$ Sorry, I should have said, "show that $f$ DOESN'T satisfy the definition from Q4"! $\endgroup$
    – Adam Saltz
    May 6, 2011 at 22:12

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