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If I have $T = (1,2]$

I want to find infimum and supremum:

$\inf(T)$ is not possible / or negative infinity

$\sup(T) = 2$

Is this correct?

It is the $\inf(T)$ that I am confused about

Thank you.

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    Check the definition of infimum and see if it fits to your answer. – 2017-02-08
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    What would you guess? – 2017-02-08
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    @copper.hat he already gave his guess in the question – 2017-02-08
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    Very loosely, think of $\inf,\sup$ as $\min, \max$. What would your guess be then? – 2017-02-08
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    @copper.hat I think you should empathize on the word $loosely$ there. – 2017-02-08
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    @theSongbird: I will try to empathize with the word loosely. – 2017-02-08
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    Thank you for your responses *she. In doing these types that I have came across, e.g. S=[1,2] I have from this that sup(S)=2, and inf(T)=1 from what you have said respectively. However the notion of the change in bracket as ( rather than [ has me somewhat perplexed. Does the change in bracket not actually mean anything? I would think logically that it would be inf(T) ≥ 1, but I am assuming that this is wrong :/ – 2017-02-09
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    @princetongirl818 sorry about that! The change in bracket does *not* affect the sup/inf, but of course it means *something*. Do you understand the difference between the sets being described? – 2017-02-09
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    dw haha. From my understanding it means that the lowest value of (1,3] for example is not 1. But [1,3] can take the lowest value of 1. – 2017-02-09
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    Right, so $(1,3] = \{x : 1 < x \leq 3\}$. Certainly, then, $(1,3]$ has a lower bound. It *doesn't* however, have a minimum (lowest element). – 2017-02-09

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$\inf(T)$ means the greatest lower-bound of $T$. We may state, for example, that all elements of $T$ are positive, so $0$ is a lower bound of $T$, which means that $\inf(T) \geq 0$. So your conclusion is clearly wrong.

In fact, we have $\inf(T) = 1$. Note that it is not necessary for $\inf(T)$ to be an element of $T$.