[Math] Real Analysis: Prove a set is closed

analysisgeneral-topologymetric-spaces

Let R be equipped with the usual Euclidean metric. Show that the set

$S=\{(x,y) \mid 0 \leq y\leq x.\}$

is closed.

The approach i am trying is to prove the complement is open, which would imply that the set is closed. I believe the complement is the union of the set where $y < 0$ and the set where $y > x$. I understand that if I can prove that if both sets are open, then the union is open. However, i get tripped up on how to prove the sets are open,

Also as a side note: i'm new here, and any help on how to use mathematical notation such as the less than or equal sign would be appreciated

Best Answer

Your approach is a good start! If you succeed in proving those two sets are open, you'll indeed be able to conclude their union is open and that their complement is closed.

Hint: To prove a set $\mathcal{O}$ is open in the ordinary Euclidean metric, you may take an arbitrary point in $\mathcal{O}$ and show that there is an open disc around it completely within $\mathcal{O}$.

An arbitrary point in the first set is an $(x,y)$ which lies below the $x$-axis. Can you find a radius around this point that stays below the $x$-axis?

Do something similar for the other set, where the point is of the form $(x,y)$ with $y>x$.

Sketching pictures and using a little simple geometry should also be helpful.

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