@JM Been there done that. The prescription free ones don't work on me and the one that apparently works (my vet friend told me) is not over the counter. So I'll just stick to Zyrtec.
@JM No I have enough of that already even without taking any pills.
@MattN Oh, sorry I missed it so far. Thank you! I know that I was being harsh and I intended to be but I certainly didn't mean to spoil an entire day... But given that I had to spend a beautiful sunny afternoon in the trunk in the corner, I guess we are even :/
@MattN: By the way: Eric's answer to that question is just an abstract version of Martin's answer, appealing to a general theorem. The proof is essentially the same.
@JM Yeah, although in recent years the pollution made the city into a urban heat whatchamccallit. It's not as humid as around the coast area but still a lot worse than a few years ago.
@Skullpatrol Can you please stop asking trivialities to particular people, rather than to the room? (I was once accused of this,not for asking trivialities though, but now, I guess I am not a defaulting bug!)
@Skullpatrol: I know where your problem come from, We have a word for this which I don't know in English. It's like CD-ROM read head sticks while reading a CD or something. I mean it gets on nerves if it takes a long time and we would turn it off.
@AsafKaragila I certainly don't like seeing threats from you to abuse the message migration tool.
(Not that there is an intended use of the message migration tool, mind you. It's just there for when you need it. I'm not sure "muting users" was what was in mind.)
@badp Yeah, that is a last resort. I already asked GraceNote to find a reasonable way to let room owners enforce some of the chat rules that were almost unanimously agreed upon.
Oh well, I'll also abuse my stars clearing powers and remove stars from my message in your general direction, because I totally see where you're coming from.
because two wrongs don't make a right but three rights make a left. Or something
@MattN: the first paragraph of 2. in your deleted answer misses the point. There need not be any injective functions on $X$ (take $X = S^1$, for example). The point is that if $g$ is not injective, say $g(x_1) = g(x_2)$ then all functions in the image of $T$ must take the same values at $x_1$, $x_2$, but not all functions on $X$ have that property (e.g. f(x) = d(x_1,x)$).
@savick01 if you add completeness, they are called proper (or spaces with the Heine-Borel property). I don't think there is a name for those spaces without completeness
@savick01 Do you mean precompact in the sense of compact closure or in the sense of compact completion? (I understood the latter for which $\mathbb{Q}$ would be an example of a space with that property that isn't locally compact).
I haven't ever heard somebody call $\mathbb{Q}$ proper.
@JM I think that I am one of the luckiest in that aspect, by the way. I write answers in a relatively esoteric field but I get a lot of votes (read: >5) on those.
@AsafKaragila I think it's just that it's easy to come up with a natural set theory question that's difficult to answer without proper knowledge. And people like learning what the answers are but don't want to learn enough set theory to be able to answer them themselves.
@JM Didier. The Watman. And my least favourite user. I don't like arrogant people. And he comes across as quite an arrogant prick. In fact, I think arrogance is a manifestation of stupidity.
And I don't quite understand your argument in point 3. The idea is that $g(X)$ is compact, so you have a point in $y \in Y \setminus g(X)$. Now consider the zero function and $d(g(X),\cdot)$ which is nonzero because $d(g(X),y) \gt 0$.
If you want $T$ to be an isometric isomorphism, then it follows that $g$ is a homeomorphism and conversely. (in fact, every isometric isomorphism comes from a homeomorphism by a theorem of Banach and Stone).
