Welcome to our community

Be a part of something great, join today!

Homeomorphism

smile

New member
Oct 15, 2013
19
Hello everyone

Here is the problem:
show that $[0,1)\times[0,1)$ is homeomorphic to $[0,1]\times[0,1)$.
I think that I need to find a continuous bijictive function between them.
however, it is hard for me to find such a function.
I am stuck on this problem, is anyone can help?

thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: homeomorphism

Hello everyone

Here is the problem:
show that $[0,1)\times[0,1)$ is homeomorphic to $[0,1]\times[0,1)$.
I think that I need to find a continuous bijictive function between them.
however, it is hard for me to find such a function.
I am stuck on this problem, is anyone can help?
I think it would be difficult to find an explicit function implementing this homeomorphism. But it is comparatively easy to see geometrically why the homeomorphism should exist.

homeo.png
Both sets $[0,1)\times[0,1)$ and $[0,1]\times[0,1)$ describe the unit square in $\mathbb R^2$, but the first one includes two sides of the boundary, while the second one contains three sides, as in the diagrams. A homeomorphism between the two sets must take the boundary of the first set to the boundary of the second set. It is obvious geometrically how to do this, since both boundaries consist of bent line segments. More explicitly, you take the boundary (the lines $AB$ and $BC$) of $[0,1)\times[0,1)$, straighten out the right-angle bend at $B$, stretch it so as to increase the length from 2 to 3, then bend it round the boundary $EFGH$ of $[0,1]\times[0,1)$. In a similar way, you can define a map from the "missing" part of the boundary of $[0,1)\times[0,1)$ ($AD$ and $DC$) to the corresponding part of the right-hand diagram (the segment $EH$). Putting all that together, you have a map defined on the boundary of $[0,1]\times[0,1]$. Call this map $f$, so that $f$ maps $AB\cup BC$ continuously to $EF\cup FG \cup GH$, and maps $AD\cup DC$ continuously to $EH$.

The next step is to extend $f$ from the boundary to a map from the entire set $[0,1]\times[0,1]$ to itself, taking $[0,1)\times[0,1)$ to $[0,1]\times[0,1)$. Again, I can see geometrically how to do this, but it would be hard to express this as an algebraic formula. In the above diagrams, the black dots show the point $\bigl(\frac12,\frac12\bigr)$. The green dots show the effect on a typical point of the boundary of $[0,1)\times[0,1)$ as the map $f$ stretches and wraps it round the boundary of $[0,1]\times[0,1)$. The idea is that if you take a point in the left diagram anywhere on the line joining the black dot and the green dot, then it should be mapped to the point that is the same fraction of the distance from the black dot to the green dot in the right diagram. So for example if the red dot in the left diagram is one-third of the way from the black dot to the green dot, then it gets mapped to the red dot in the right diagram, which is also one-third of the way from the black dot to the green dot.

Trying to describe that map in words was quite cumbersome, but the basic idea is simple, namely to pivot the set $[0,1)\times[0,1)$ around its midpoint, keeping the top left corner fixed, stretching the lower left half of the square and shrinking the upper right half, so as to twist it to look like the set $[0,1]\times[0,1)$. As I said at the start, I think it would be difficult to describe such a map algebraically.