Hi I am wondering if f(x)=2 is continuous for $\displaystyle x\in [0,3)\cup \{4\}$?

I'd imagine f(x)=2 would be continuous on any domain, just not too sure.

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- Oct 26th 2009, 07:20 PMdannyboycurtisis f(x)=2 continuous for the given domain?
Hi I am wondering if f(x)=2 is continuous for $\displaystyle x\in [0,3)\cup \{4\}$?

I'd imagine f(x)=2 would be continuous on any domain, just not too sure. - Oct 26th 2009, 07:25 PMMush
- Oct 26th 2009, 07:52 PMtonio

Well, $\displaystyle [0,3)\cup {4}$ is not a domain because it is not an open set (I'm assuming we're talking about the real line with the usual euclidean topology).

In this case the question "is f(x) continuous at x=4?" is meaningless since there exists an open neighborhood of 4 where f(x) is not defined and thus limits processes, which are needed for continuity, cannot be carried on with the point 4.

Tonio - Oct 26th 2009, 07:55 PMdannyboycurtis
but because x exists in [0,3) or x exists in {4},

doesnt it suffice to show that f(x)=2 is continuous in [0,3). And if so, would it be continous in that interval?? - Oct 26th 2009, 08:05 PMtonio

Yes, on that INTERVAL f is continuous, but you cannot talk of conitnuity in an isolated point, meaning: a point which has some open neighborhood around it where the function isn't defined.

As continuity is defined by means of limits we HAVE to have the possibility to evaluate limits. In your example, with x = 4 we can't do that.

Tonio - Oct 26th 2009, 08:08 PMdannyboycurtis
ok, so as I now understand it,

f(x) =2 is discontinuous at {4} because there is no neighborhood around 4 to evaluate the limit with.

but f(x)=2 IS continuous on [0,3)U{4} because f(x) is continous on [0,3)

sorry to beat a dead horse but I just to make sure I understand what you are saying... - Oct 26th 2009, 08:12 PMtonio
- Oct 26th 2009, 08:14 PMdannyboycurtis
but doesnt the definition of union mean that x exists in [0,3) OR {4}?

- Oct 26th 2009, 08:15 PMdannyboycurtis
oh I see nevermind, thanks for the help tonio!

- Oct 26th 2009, 08:20 PMMush
In order for a function to be continuous on some interval, then there must be a neighbourhood around each point in the interval about which limits can be taken which also lie in the interval.

In your interval, the element 4 does not have a neighbourhood which is also in the interval, so limits cannot be taken about the point x = 4 for the function f(x), because f(x) is not defined on the neighbourhood of x = 4.

Continuity is a concept that is defined by limits. If a function is continuous, then a forward and backward limit must exist at EVERY POINT in the intervals for which the function is defined.

In your interval, the function f(x) has no limit whatsoever at the point x = 4, because it is not defined on the neighbourhood of 4.

So f(x) is continuous on [0,3), but is it not continuous on [0,3) U {4}.

Mathematically speaking, a function is continuous in the neighbour hood of a point, a if:

$\displaystyle \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x)$.

In your case, the function does not 'approach' 4 through either negative or positive values. It doesn't get 'approached' at all. And hence the limits do not exist, and continuity around the neighbourhood of x = 4 cannot be established because said neighbourhood is nonexistent.