# Examples illustrating the difference between closed and bounded sets.

Intuitively for me, it seems as if closed sets are bounded, especially considering closed sets contain all limit points. But I know this isn't the case, because $$ℝ$$ is closed (and open) and is not bounded.

Is this the only case of a closed set not being bounded? Can anyone provide an example that further illustrates the difference between closed and bounded?

• One thing it might be handy to remember is that a closed set is the complement of an open set. So like, the complement of any open ball, for instance, is closed.
– crf
Commented Dec 8, 2012 at 8:19
• An easier way to describe bounded (for me) is that the set lies in a closed ball. Commented Jul 24, 2022 at 10:14

## 4 Answers

We cover each of the four possibilities below.

Closed and bounded: $[0,1]$

Closed and not bounded: $\cup_{n\in Z}[2n,2n+1]$

Bounded and not closed: $(0,1)$

Not closed and not bounded: $\cup_{n\in Z}(2n,2n+1)$

The integers as a subset of $\Bbb R$ are closed but not bounded.

• Do you mean ℤ? So, would every finite set be closed and bounded?
– Alti
Commented Dec 8, 2012 at 0:24
• yes this is because each singleton is closed and the finite union of closed sets is closed again
– Amr
Commented Dec 8, 2012 at 0:26
• Finally finite implies bounded
– Amr
Commented Dec 8, 2012 at 0:27
• @sugataAdhya I was referring to singletons in $\mathbb{R}$ not in any space
– Amr
Commented Feb 10, 2014 at 16:28

$$\{x\in\mathbb R\mid x\geq 0\}$$

Also note that there are bounded sets which are not closed, for examples $\mathbb Q\cap[0,1]$.

• Thanks, can you please see my reply to Amr?
– Alti
Commented Dec 8, 2012 at 0:25
• Yes, every finite set is closed and definitely bounded. Commented Dec 8, 2012 at 0:26
• @AsafKaragila: Check out my comment above. Commented Dec 8, 2012 at 8:16
• @Sugata: Real analysis is done in the real numbers with the standard topology. Commented Dec 8, 2012 at 8:45
• @AsafKaragila: 'General-Topology' is tagged with the question. But as far as your solution is concerned it's all right. Commented Dec 8, 2012 at 9:09

In $\mathbb R^n$ every non-compact closed set is unbounded.