Let $a$ and $b$ be arbitrary real numbers with $a < b$. Show that $[a,b]$ is closed by proving its complement is open.

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I don't have any idea on this, can anyone help me on this?

Write down what it means for a set to be open. Show that $[a,b]^C$ satisfies this condition. — David Mitra, Jan 25 '15 at 11:49
Yes, anymore than that hint would be akin to giving the solution outright. It may be safe to add: write out the complement as a union of certain sets, then note that they are both open, and conclude the union is open (why?). Best of luck! — Musa Al-hassy, Jan 25 '15 at 11:52

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

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The answer was essentially given on comments. Here are some references.

Any open interval is an open set. Also, the union of open sets is open.

The complement of $[a,b]$ is $(-\infty,a)\cup(b,\infty)$. So, it's open (union of open sets).

edited Apr 13 '17 at 12:21

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answered Jan 25 '15 at 12:08

Pedro

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score 1 · Accepted Answer · answered Jan 26 '15 at 00:02

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Using the definition of openness that was given in your previous question. Note that the complement of $[a,b]$ is $]\infty,a[ \cup ]b,\infty[$. For any point in this set, you can find a small open ball that is contained within the set.

answered Jan 26 '15 at 00:02

Loreno Heer

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Let $a$ and $b$ be arbitrary real numbers with $a < b$. Show that $[a,b]$ is closed by proving its complement is open.

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