Show, from the definitions of open and closed sets, that when using the standard Euclidean metric, [0, 1) × [0, 1) is neither an open nor closed subset of $\Bbb{R^2}$.

From what I understand, a set is open if every point in the set has a neighbourhood contained within the set. Closed is the complement of an open set.

Since (0,0) is an element of the set that lies on the boundary, it implies that the set is not open as there exist points in the neighbourhood of (0,0) which are not elements of the set.

Since all limit points (e.g.(1,1)) are not included in the set, the set is not closed.

Therefore, the set is neither open nor closed.

But, how do I use the standard Euclidean metric to show that? This is where I'm stuck. Please help.

What do I do next?

  • 1
    $\begingroup$ What are your definitions of "open" and "closed"? $\endgroup$ – Adam Hughes Mar 25 '15 at 4:54
  • $\begingroup$ A set is open if every point in the set has a neighbourhood contained within the set. Closed is the complement of an open set. $\endgroup$ – user117239 Mar 25 '15 at 5:05

Hint: It doesn't have a neighbourhood for $(0,0)$ and its complement doesn't have a neighbourhood for $(1,1)$

  • $\begingroup$ I edited the question to reflect my understanding. Would you be able to help now? $\endgroup$ – user117239 Mar 25 '15 at 5:16

Define $X$ to be the square $[0,1) \times [0,1),$ viewed as a subset of $\mathbb{R} \times \mathbb{R}$.

Its probably best to start out by defining $X$ very explicitly, in low level "machine code." So you would write:

Then for all $(x,y) \in \mathbb{R} \times \mathbb{R},$ the following are equivalent.

  1. $(x,y) \in X$
  2. $0 \leq x < 1$, and $0 \leq y < 1$.

So anywhere you see something of the general form $(\mathrm{stuff}_0,\mathrm{stuff}_1) \in X$, you can replace it with the appropriate version of (2), and anytime you see a version of (2), you can replace with with $(\mathrm{stuff}_0,\mathrm{stuff}_1) \in X$ for appropriate choices of expressions $\mathrm{stuff}_0$ and $\mathrm{stuff}_1$.


To show that $X$ isn't open, consider the point $(0,0)$. Now assume for a contradiction that some neighbourhood $N$ of $(0,0)$ is a subset of the square $X$. Since $N$ is a neighbourhood of $(0,0),$ there exists a real number $r > 0$ such that $B_r(0,0) \subseteq N$. So $B_r(0,0) \subseteq X$. Can you see how to derive a contradiction from here? Let me give you a hint: the ordered pair $(-r/2,-r/2)$ is an element of $B_r(0,0)$. We're almost there, so I'll leave the rest to you. Remember, you're looking for a contradiction, so you're trying to write down the negation of (any!) statement that you've already written down.

Essentially the same method should work for for showing that the complement of $X$ isn't open, except that instead of $(-r/2,-r/2)$ you'll have to choose a different point. Draw a picture and it should be obvious which point to take. I also suggest writing a very explicit definition of $X^c$, just like we did with $X$; experience tells me this often helps somewhat.

Good luck!


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