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Let $x \in X$ and define a topological space $(X, \tau)$ and let singleton set {$x$} $\in \tau$. Then by definition of neighborhood of a point in topology, {${x}$} will be a neighborhood of point $x$. My question is

If set {$x$} does not contain any other point then $x$, then how does it make sense to say that this singleton set is a neighborhood of $x$.

Andrew Miller
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  • What topology are you considering here? – Mikasa Jan 19 '14 at 20:18
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    It's because you are defining the singleton set to be an open set in your topology. – monroej Jan 19 '14 at 20:18
  • @monroej by definition it is true, but neighborhood of $x$ in real analysis contains some points around $x$, here there is no other point. – Andrew Miller Jan 19 '14 at 20:21
  • @B.S. let say topology $\tau$ is the whole power set. – Andrew Miller Jan 19 '14 at 20:22
  • @AbhinavSahani True... but you aren't using either the real numbers or their usual topology here. So, the "shape" of the space can be very different! Saying that ${x}$ is open can be thought of as saying that there are "no other points near $x$". – Nick Peterson Jan 19 '14 at 20:22
  • In topology a neighborhood of a point $x$ is just a set $V$ that contains an open set $U$ with $x\in U$. But I totally see where the confusion arises! – monroej Jan 19 '14 at 20:22
  • To see a case in which this ''really'' happens: look at $X = [0,1] \cup {2}$ with the topology induced by $\mathbb R$. ${2}$ is indeed a neighborhood of the point 2 - and that doesn't look at all unreasonable, does it? – Magdiragdag Jan 19 '14 at 20:22
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    So it's a neighborhood of $x$ that does not really contain any 'neighbours' of $x$. Mathematicians are kind of peculiar, aren't they :)-. – drhab Jan 19 '14 at 20:23
  • One well-known possibility is to consider Discrete Metric on your non empty set $X$. – Mikasa Jan 19 '14 at 20:24
  • @NicholasR.Peterson that is quite ok. but let us consider the topology to be whole power set, there we may find neighborhoods that will have some points around $x$ and also {$x$}, then how will your sentence make sense? – Andrew Miller Jan 19 '14 at 20:29
  • The fact that there ARE open sets which include $x$ and other points doesn't change the fact that you've declared ${x}$ an open set. Because ${x}$ is itself open, you can separate $x$ from all other points in $X$. – Nick Peterson Jan 19 '14 at 20:30
  • @monroej doesn't neighborhood have to be open? – Andrew Miller Jan 19 '14 at 20:34
  • It depends on your definition. Some people say "open neighborhood" of $x$ to mean...well just that, an open set containing $x$. Other people say neighborhood to mean open neighborhood. It really just depends on your text. Luckily, in this case, it didn't matter. – monroej Jan 19 '14 at 20:36
  • @NicholasR.Peterson so can we say that a isolated point $x$ is a neighborhood of that point?(talking in the sense of real analysis) – Andrew Miller Jan 19 '14 at 20:37
  • I like the definition of neighborhood that I put in my second comment. But that's just me. – monroej Jan 19 '14 at 20:37
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    A point $x$ is isolated if and only if ${x}$ is open in the topology, yes. – Nick Peterson Jan 19 '14 at 20:38
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    @NicholasR.Peterson beat me to it :) – monroej Jan 19 '14 at 20:39

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You stated that the singleton is an open set (you said {$x$} $\in \tau$). An open set is a neighbourhood of any of its own elements, by the definition of neighbourhood.

ir7
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