Is the graph of f(x)=1/x a closed set?

[yifli]

f(x)=1/x closed set??

A book I'm reading now says the graph of f(x)=1/x is a closed set, how come??

Its range is [(-\infty,0)\cup (0, \infty). A set is closed iff every convergent sequence has a limit point in the set. If a sequence converges to 0, then 0 is not in the range

 

[spamiam]

I think you're forgetting part of your definition for closed.

A set S \subseteq \mathbb{R}^2 is closed if it contains all of its limit points, i.e. if every convergent sequence contained in S converges to a point in S. There are no sequences contained in the graph of f(x) = 1/x that converge to 0.

An alternative definition for closed may make it easier to see that this set is closed. A set is closed if and only if its complement is open. The complement of the graph is certainly open, since for any point not contained in the graph, we can find an open disc containing the point that doesn't intersect the graph.

 

[dextercioby]

The graph is the subset of mathbb{R}\oplus mathbb{R}

\mathcal{G} = \left\{\mathbb{R}-\{0}, \mathbb{R}-\{0\}\right\}

This is closed, since all sequences from this set converge to elements of this set in the metric topology.

 

[yifli]

I think you're forgetting part of your definition for closed.

A set S \subseteq \mathbb{R} is closed if it contains all of its limit points, i.e. if every convergent sequence contained in S converges to a point in S. There are no sequences contained in the graph of f(x) = 1/x that converge to 0.

what about sequence {1/n}?

 

[spamiam]

As dextercioby pointed out, and as I should have, the graph is actually a subset of \mathbb{R}^2, so I think your sequence would have to be {(n, 1/n)}, which doesn't converge, since the first coordinate goes to infinity.